Laws of Motion

Concept. A constant force gives a constant acceleration $\vec a=\vec F/m$. From rest at the origin the position is $\vec r=\tfrac12\vec a\,t^2$, and each component can be handled on its own.

Step 1. Find the acceleration. Convert the mass: $m=100$ g $=0.1$ kg. $$\vec a=\frac{\vec F}{m}=\frac{5\hat i+10\hat j}{0.1}=50\hat i+100\hat j\ \text{m s}^{-2}$$

Step 2. Position after 2 s. The body starts from rest ($\vec u=0$), so $$\vec r=\tfrac12\vec a\,t^2=\tfrac12(50\hat i+100\hat j)(2)^2=100\hat i+200\hat j\ \text{m}$$

Step 3. Compare with the given form. Match the components with $2x\hat i+5y\hat j$: $$2x=100\Rightarrow x=50,\qquad 5y=200\Rightarrow y=40$$

Answer. $x:y=50:40=5:4$, option D. Note that $x$ and $y$ are not the coordinates themselves; they are defined through $2x$ and $5y$.

Concept. By Newton's second law $F=ma$. When the velocity is given as a function of position (not time), the acceleration is found from the chain rule $a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v\dfrac{dv}{dx}$.

Step 1. Differentiate $v$ with respect to $x$. $$v=10\sqrt x\ \Rightarrow\ \frac{dv}{dx}=\frac{10}{2\sqrt x}=\frac{5}{\sqrt x}$$

Step 2. Acceleration. $$a=v\frac{dv}{dx}=10\sqrt x\cdot\frac{5}{\sqrt x}=50\ \text{m s}^{-2}$$ The $\sqrt x$ cancels, so the acceleration is constant.

Step 3. Force. With $m=500$ g $=0.5$ kg, $$F=ma=0.5\times50=25\ \text{N}$$

Answer. $25$ N, option B. Shortcut: $v=k\sqrt x$ is the same as $v^2=k^2x$, which is $v^2=2ax$ with $a=k^2/2$.

Concept. Isolate the monkey. Only two forces act on it: its weight $mg$ down and the rope's tension $T$ up. Newton's second law then gives $T$, and the sign of the acceleration (not the direction of motion) decides whether $T$ is more or less than $mg$.

Step 1. Climbing down with acceleration $4$ m s$^{-2}$ downward. Take downward as positive: $$mg-T=ma\ \Rightarrow\ T=m(g-a)=50(10-4)=300\ \text{N}$$ This is below the $350$ N limit, so the rope holds.

Step 2. Climbing up with acceleration $5$ m s$^{-2}$ upward. Take upward as positive: $$T-mg=ma\ \Rightarrow\ T=m(g+a)=50(10+5)=750\ \text{N}$$

Step 3. Compare with the breaking load. $750\ \text{N}>350\ \text{N}$, so the rope cannot supply this tension.

Answer. The rope breaks while the monkey climbs upward, option C. Options A and B quote tensions that are not what the rope would need ($750$ N and $300$ N).

Concept. Two forces act on the balloon system: the upthrust $U$ of the air (it depends only on the balloon's volume, so it does not change when ballast is dropped) and the weight. Write Newton's second law before and after releasing mass, then eliminate $U$.

Step 1. Before release. Mass $M$, acceleration $a$ upward: $$U-Mg=Ma\ \Rightarrow\ U=M(g+a)$$

Step 2. After releasing mass $m$. The remaining mass $M-m$ has acceleration $3a$ upward, with the same $U$: $$U-(M-m)g=(M-m)\,3a\ \Rightarrow\ U=(M-m)(g+3a)$$

Step 3. Equate the two expressions for $U$. $$M(g+a)=(M-m)(g+3a)$$ $$Mg+Ma=Mg+3Ma-m(g+3a)\ \Rightarrow\ m(g+3a)=2Ma$$ $$m=\frac{2Ma}{3a+g}$$

Answer. Option A. Check: if the new acceleration were still $a$, the formula $\dfrac{M(a'-a)}{g+a'}$ gives $m=0$, as it should.

Concept. Air resistance always opposes the velocity. Going up, it acts downward together with gravity; coming down, it acts upward against gravity. So the two accelerations differ, and the same height is covered in different times.

Step 1. Acceleration due to air resistance. $$\frac{f}{m}=\frac{10}{5}=2\ \text{m s}^{-2}$$

Step 2. Ascent. Weight and drag both point down: $$a_u=g+\frac fm=10+2=12\ \text{m s}^{-2}\ (\text{retardation})$$

Step 3. Descent. Weight down, drag up: $$a_d=g-\frac fm=10-2=8\ \text{m s}^{-2}$$

Step 4. Relate time to acceleration. The ascent ends at rest at height $h$ and the descent starts from rest at $h$. Reversing the ascent in time, both are motions from rest over $h$, so $h=\tfrac12at^2$ and $t=\sqrt{2h/a}\propto\dfrac{1}{\sqrt a}$. $$\frac{t_u}{t_d}=\sqrt{\frac{a_d}{a_u}}=\sqrt{\frac{8}{12}}=\sqrt{\frac23}=\frac{\sqrt2}{\sqrt3}$$

Answer. $\sqrt2:\sqrt3$, option B. With any resistance the ascent is quicker than the descent, so options with ratio $\ge1$ can be rejected at once.

Concept. The only force along the motion is the resistance $F=-kv^2$, so Newton's second law gives a differential equation $m\dfrac{dv}{dt}=-kv^2$. Separate the variables and integrate. The kinetic energy condition gives the final speed.

Step 1. Speed after 10 s. $$\tfrac12mv^2=\tfrac18mv_0^2\ \Rightarrow\ v^2=\frac{v_0^2}{4}\ \Rightarrow\ v=\frac{v_0}{2}=5\ \text{m s}^{-1}$$

Step 2. Set up and separate. $$m\frac{dv}{dt}=-kv^2\ \Rightarrow\ \frac{dv}{v^2}=-\frac km\,dt$$

Step 3. Integrate from $v_0$ at $t=0$ to $v$ at $t$. $$\int_{v_0}^{v}\frac{dv}{v^2}=-\frac km\int_0^t dt\ \Rightarrow\ -\frac1v+\frac1{v_0}=-\frac km t\ \Rightarrow\ \frac1v-\frac1{v_0}=\frac km t$$

Step 4. Substitute. $m=10^{-2}$ kg, $t=10$ s: $$\frac15-\frac1{10}=\frac{k}{10^{-2}}\times10\ \Rightarrow\ 0.1=1000\,k\ \Rightarrow\ k=10^{-4}$$

Step 5. Unit. $kv^2$ is a force, so $[k]=\dfrac{\text{kg m s}^{-2}}{\text{m}^2\text{s}^{-2}}=\text{kg m}^{-1}$.

Answer. $k=10^{-4}$ kg m$^{-1}$, option D. The unit alone rules out options A and C.

Concept. Newton's second law in its original form is $F=\dfrac{dp}{dt}$. On a $p$–$t$ graph this is the slope, so the steepest region has the largest force and the flattest region the smallest.

Step 1. Read the regions. In the figure, region a (from $0$ to $t_1$) has $p$ rising moderately; region b (from $t_1$ to $t_2$) has $p$ rising only slightly over a long time; region c (from $t_2$ to $t_3$) has $p$ falling all the way to zero.

Step 2. Compare slopes.

• Region c: the largest change in momentum (from its peak to zero) over the shortest time, since $t_3-t_2<t_1$. Steepest slope.
• Region a: a smaller change over a time $t_1$. Moderate slope.
• Region b: the smallest rise over the longest interval. Flattest slope.

Step 3. Convert to force. $|F|=\left|\dfrac{\Delta p}{\Delta t}\right|$ is largest in c and smallest in b. The sign of the slope only gives the direction of the force; the question asks for magnitude.

Answer. Maximum in c, minimum in b: option C.

Concept. By Newton's second law the force on the bullets equals the momentum given to them per second. By the third law the bullets push back on the gun with the same force, and the person holding the gun must supply it.

Step 1. Momentum of one bullet. $$p=mv=0.01\times250=2.5\ \text{kg m s}^{-1}$$

Step 2. Momentum per second. If $n$ bullets are fired per second, the momentum given per second is $n\,mv$, so $$F=n\,mv$$

Step 3. Solve for $n$. $$n=\frac{F}{mv}=\frac{125}{2.5}=50\ \text{per second}$$

Answer. $50$, option A. Remember to convert $10$ g to $0.01$ kg; leaving it in grams gives $0.05$.

Concept. Impulse equals the change in momentum, $\vec J=\Delta\vec p$. The body arrives moving down and leaves moving up, so the two momenta are in opposite directions and their magnitudes add.

Step 1. Speed just before hitting the ground. Free fall from $h_1=10$ m: $$v_1=\sqrt{2gh_1}=\sqrt{2\times9.8\times10}=\sqrt{196}=14\ \text{m s}^{-1}\ (\text{down})$$

Step 2. Speed just after the bounce. It rises to $h_2=5$ m, so $$v_2=\sqrt{2gh_2}=\sqrt{98}\approx9.9\ \text{m s}^{-1}\ (\text{up})$$

Step 3. Change in momentum. Take upward as positive: $p_i=-mv_1$, $p_f=+mv_2$. $$J=p_f-p_i=m(v_2+v_1)=0.1\times(9.9+14)=0.1\times23.9\approx2.39\ \text{kg m s}^{-1}$$

Answer. $2.39$ kg m s$^{-1}$, upward, option D. Subtracting the speeds ($0.41$) is the usual mistake: momentum is a vector.

Concept. For a force that varies with time, the impulse is $J=\int F\,dt$, and it equals the change in momentum. The particle starts from rest, so its final momentum is $Mv=J$.

Step 1. Write the impulse. $$J=\int_0^{2T}F_0\left[1-\left(\frac{t-T}{T}\right)^2\right]dt$$

Step 2. Substitute to simplify. Let $u=\dfrac{t-T}{T}$, so $dt=T\,du$. When $t=0$, $u=-1$; when $t=2T$, $u=1$. $$J=F_0T\int_{-1}^{1}(1-u^2)\,du=F_0T\left[u-\frac{u^3}{3}\right]_{-1}^{1}=F_0T\left(\frac23+\frac23\right)=\frac{4F_0T}{3}$$

Step 3. Final velocity. The direction of the force is fixed, so $$Mv=J\ \Rightarrow\ v=\frac{4F_0T}{3M}$$

Answer. Option C. Check: the force curve is a parabola with peak $F_0$ over a base $2T$; the area of such a parabolic arch is $\tfrac23\times$ base $\times$ height $=\tfrac23(2T)F_0$.

Concept. The force on the mesh equals the rate of change of momentum of the liquid that hits it. Different fractions of the liquid have different momentum changes, so add their contributions. Pressure is then force divided by area.

Step 1. Mass arriving per second. In one second a column of length $v$ and cross-section $A=\pi a^2$ leaves the pipe: $$\frac{dm}{dt}=\rho Av$$

Step 2. Momentum change per unit mass for each part. Each kilogram arrives with momentum $v$.

• $50\%$ passes through unaffected: change $0$.
• $25\%$ stops: change $v$.
• $25\%$ comes back with speed $v$: change $v-(-v)=2v$.

Step 3. Force on the mesh. $$F=0.25\,\rho Av\cdot v+0.25\,\rho Av\cdot2v=0.75\,\rho Av^2=\frac34\rho Av^2$$

Step 4. Pressure. $$P=\frac FA=\frac34\rho v^2$$

Answer. Option B. A rebounding fraction counts twice as much as a stopped fraction.

Concept. Sand lands on the belt with no horizontal velocity and must be brought up to the belt's speed $v$. The momentum given to the sand per second is $v\dfrac{dm}{dt}$, so this is the extra force the motor must supply to keep $v$ constant. The power is $P=Fv$.

Step 1. Force needed. $$F=\frac{d(mv)}{dt}=v\frac{dm}{dt}\quad(v\text{ constant})$$

Step 2. Use the given rate. $\dfrac{dm}{dt}\propto\sqrt v$, so $$F\propto v\cdot v^{1/2}=v^{3/2}$$

Step 3. Power. $$P=Fv\propto v^{3/2}\cdot v=v^{5/2}$$

Step 4. Match the options. Squaring removes the fraction: $$P^2\propto v^5$$

Answer. Option C. Do not use $\tfrac12v^2\dfrac{dm}{dt}$ (the rate of gain of kinetic energy) for the power: the other half of $P$ is lost as heat while the sand slips on the belt.

Concept. The exhaust gases leave at speed $u$ relative to the rocket, so they push the rocket forward with a thrust $u\dfrac{dm}{dt}$ (rate of momentum given to the gas). Isolate the rocket: thrust up, weight down, and apply Newton's second law with the current mass.

Step 1. Forces on the rocket at the start. Mass $m=1000$ kg, upward thrust $u\dfrac{dm}{dt}$, weight $mg$ down.

Step 2. Newton's second law (upward positive). $$u\frac{dm}{dt}-mg=ma$$

Step 3. Solve for the burn rate. $$\frac{dm}{dt}=\frac{m(g+a)}{u}=\frac{1000\times(10+20)}{500}=60\ \text{kg s}^{-1}$$

Answer. $60$ kg s$^{-1}$, option B. Forgetting gravity gives $40$ kg s$^{-1}$, which is not offered, a hint that $g$ must be included.

Concept. The body is in equilibrium under three concurrent forces: $T_1$, $T_2$ and $mg$. Resolve them horizontally and vertically and set each sum to zero.

Step 1. Read the angles. From the figure, $T_1$ (left string) makes $60^\circ$ with the horizontal and $T_2$ (right string) makes $30^\circ$. The angle between the strings is $180^\circ-60^\circ-30^\circ=90^\circ$, so they are perpendicular.

Step 2. Horizontal balance. $$T_1\cos60^\circ=T_2\cos30^\circ\ \Rightarrow\ \frac{T_1}{2}=\frac{\sqrt3\,T_2}{2}\ \Rightarrow\ T_1=\sqrt3\,T_2$$

Step 3. Vertical balance. $mg=1\times10=10$ N: $$T_1\sin60^\circ+T_2\sin30^\circ=10$$ $$\sqrt3T_2\cdot\frac{\sqrt3}{2}+\frac{T_2}{2}=\frac{3T_2}{2}+\frac{T_2}{2}=2T_2=10\ \Rightarrow\ T_2=5\ \text{N}$$

Step 4. Back-substitute. $T_1=\sqrt3\times5=5\sqrt3$ N.

Answer. $5\sqrt3$ N and $5$ N, option B. Shortcut for perpendicular strings: each tension is $mg\cos(\text{its angle with the vertical})$: $T_1=10\cos30^\circ$, $T_2=10\cos60^\circ$. The steeper string carries more.

Concept. The block is at rest on a smooth incline, so only three forces act: weight $mg$, normal force $N$ (perpendicular to the incline) and the horizontal force $F$. Resolving along the incline removes $N$ and gives $F$ directly.

Step 1. Forces and axis. The incline makes $60^\circ$ with the horizontal and $F$ pushes horizontally towards the incline. Take the axis along the incline.

• Weight: component $mg\sin60^\circ$ down the incline.
• $F$: component $F\cos60^\circ$ up the incline (the horizontal makes $60^\circ$ with the incline).
• $N$: no component along the incline.

Step 2. Balance along the incline. $$F\cos60^\circ=mg\sin60^\circ\ \Rightarrow\ F=mg\tan60^\circ$$

Step 3. Substitute. $m=0.2$ kg, $g=10$ m s$^{-2}$: $$F=0.2\times10\times\sqrt3=2\sqrt3=\sqrt{12}\ \text{N}$$

Answer. $x=12$. On a smooth incline this horizontal force is the only one that holds the block, so "minimum" just fixes the case; the normal force is $mg\sec60^\circ=4$ N.

Concept. For a point in equilibrium, the components of all forces along each axis add to zero. Choose $x$ to the right (along $F_1$) and $y$ upward (opposite to $F_2$).

Step 1. Read the figure. At $P$: $F_1$ acts to the right, $F_2$ acts downward, the $2$ N force points up and to the left at $45^\circ$ to the vertical, and the $1$ N force points up and to the right at $45^\circ$ to the vertical. So each slanted force has components of magnitude $\text{force}\times\dfrac1{\sqrt2}$ along both axes.

Step 2. Sum of $x$-components. $$F_1+\frac{1}{\sqrt2}-\frac{2}{\sqrt2}=0\ \Rightarrow\ F_1=\frac{1}{\sqrt2}\ \text{N}$$

Step 3. Sum of $y$-components. $$\frac{1}{\sqrt2}+\frac{2}{\sqrt2}-F_2=0\ \Rightarrow\ F_2=\frac{3}{\sqrt2}\ \text{N}$$

Step 4. Ratio. $$F_1:F_2=\frac{1}{\sqrt2}:\frac{3}{\sqrt2}=1:3$$

Answer. $x=3$. The $\sqrt2$ cancels, so the ratio is just $(2-1):(2+1)$.

Concept. Treat the rope as light. Then the lower half simply transmits the weight, and the point where the horizontal force acts is in equilibrium under three forces. The upper half's tension must balance the resultant of the other two.

Step 1. Lower half. It holds the $10$ kg mass at rest, so it is vertical and its tension is $$T_2=Mg=10\times10=100\ \text{N}$$ It pulls the midpoint straight down with $100$ N.

Step 2. Forces at the midpoint. Three forces: $100$ N down (lower half), $30$ N horizontal (applied), and $T_1$ along the upper half, at angle $\alpha$ to the vertical.

Step 3. Resolve. $$T_1\sin\alpha=30,\qquad T_1\cos\alpha=100$$ Dividing, $$\tan\alpha=\frac{30}{100}=0.3=3\times10^{-1}$$

Answer. $x=3$. The rope length ($5$ m) and the position of the point on the rope do not matter for a light rope.

Concept. A hanging chain is in equilibrium, so the forces on any part of it balance. Take the whole chain to find the tension at the supports, then take half the chain: at the lowest point the chain is horizontal, so the tension there is horizontal.

Step 1. Whole chain. At each support the tension $T$ acts along the chain, at $30^\circ$ to the horizontal. By symmetry both are equal. The vertical components carry the weight: $$2T\sin30^\circ=mg\ \Rightarrow\ T=mg$$

Step 2. One half of the chain. Forces on it: $T$ at the support (at $30^\circ$ above the horizontal), the horizontal tension $T_0$ at the lowest point, and its weight $mg/2$ (vertical).

Step 3. Horizontal balance on the half. Only $T$ and $T_0$ have horizontal parts: $$T_0=T\cos30^\circ=mg\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{2}mg$$ (Check vertically: $T\sin30^\circ=\dfrac{mg}{2}$, which equals the half's weight.)

Answer. $\dfrac{\sqrt3}{2}mg$, option B. In general $T_0=\dfrac{mg}{2}\cot\theta$.

Concept. The strings are taut, so all three blocks share one acceleration. Find it from the whole system, then isolate pieces: a string's tension is the only horizontal force on the blocks behind it, so it equals their total mass times $a$.

Step 1. Common acceleration. The surface is smooth, and the only external horizontal force is $F=80$ N: $$a=\frac{F}{m_A+m_B+m_C}=\frac{80}{5+3+2}=8\ \text{m s}^{-2}$$

Step 2. Tension $T_1$ (between $A$ and $B$). Isolate $A$. The only horizontal force on it is $T_1$: $$T_1=m_Aa=5\times8=40\ \text{N}$$

Step 3. Tension $T_2$ (between $B$ and $C$). Isolate $A$ and $B$ together. The only external horizontal force on them is $T_2$: $$T_2=(m_A+m_B)a=8\times8=64\ \text{N}$$

Step 4. Check with $C$. $F-T_2=m_Ca$: $80-64=16=2\times8$. Correct.

Answer. $40$ N and $64$ N, option A. Tension grows towards the pulling end, since each string drags more mass.

Concept. A spring pulls its two ends with equal and opposite forces. Block $A$ has only the spring acting on it horizontally, so its acceleration tells us the spring force. That same force then acts on $B$ in the opposite direction.

Step 1. Isolate $A$. Only the spring force $F_s$ acts horizontally. Its acceleration has magnitude $a$, so $$F_s=Ma$$ $A$ is being pulled towards $B$ (the spring is stretched as $B$ is pulled away by $F$).

Step 2. Isolate $B$. Two horizontal forces: the applied force $F$, directed away from $A$, and the spring force $Ma$, directed towards $A$.

Step 3. Newton's second law for $B$ (positive in the direction of $F$): $$F-Ma=Ma_B\ \Rightarrow\ a_B=\frac{F-Ma}{M}$$

Answer. Option D. Check with the whole system: the spring forces are internal, so $F=Ma+Ma_B$, which gives the same $a_B$. (The question's words "moving away" are loose; the spring is stretched, so $A$ is actually dragged along after $B$. Only the size $a$ of $A$'s acceleration matters.)

Concept. In an Atwood machine the heavier mass $m_1$ goes down and $m_2$ goes up with the same acceleration (one light string). Write Newton's law for each mass and add to eliminate the tension.

Step 1. Equations of motion. $$m_1:\ m_1g-T=m_1a,\qquad m_2:\ T-m_2g=m_2a$$

Step 2. Add them. $T$ cancels: $$(m_1-m_2)g=(m_1+m_2)a\ \Rightarrow\ a=\frac{(m_1-m_2)g}{m_1+m_2}$$

Step 3. Use $a=g/8$. $$\frac{m_1-m_2}{m_1+m_2}=\frac18\ \Rightarrow\ 8m_1-8m_2=m_1+m_2\ \Rightarrow\ 7m_1=9m_2$$

Answer. $\dfrac{m_1}{m_2}=\dfrac97$, option A. General result: $a=g/n$ gives $\dfrac{m_1}{m_2}=\dfrac{n+1}{n-1}$.

Concept. All three masses move with the same acceleration magnitude: $m_1$ on one side, $m_2$ and $m_3$ (hanging one below the other) on the other. Find $a$ from the whole system, then isolate single masses to get each tension.

Step 1. Read the figure. $T_1$ is the tension in the string over the pulley (it holds $m_1$ on the left and $m_2$ on the right). $T_2$ is the tension in the short string between $m_2$ and $m_3$.

Step 2. Acceleration. Left side $4$ kg, right side $m_2+m_3=10$ kg, so the right side goes down: $$a=\frac{(10-4)g}{4+10}=\frac{60}{14}=\frac{30}{7}\ \text{m s}^{-2}$$

Step 3. Isolate $m_1$ (accelerating up). Forces: $T_1$ up, $m_1g$ down: $$T_1-m_1g=m_1a\ \Rightarrow\ T_1=4\left(10+\frac{30}{7}\right)=4\times\frac{100}{7}=\frac{400}{7}\ \text{N}$$

Step 4. Isolate $m_3$ (accelerating down). Forces: $T_2$ up, $m_3g$ down: $$m_3g-T_2=m_3a\ \Rightarrow\ T_2=6\left(10-\frac{30}{7}\right)=6\times\frac{40}{7}=\frac{240}{7}\ \text{N}$$

Step 5. Ratio. $$\frac{T_1}{T_2}=\frac{400}{240}=\frac53$$

Answer. Option A. Check with $m_2$ (moving down, pulled up by $T_1$ and down by $T_2$ and its weight): $m_2g+T_2-T_1=40+\dfrac{240-400}{7}=\dfrac{120}{7}$, and $m_2a=4\times\dfrac{30}{7}=\dfrac{120}{7}$. Correct.

Concept. The balls form one connected system with a common acceleration. On a smooth table, the only driving force is the weight of the hanging balls. A string between two balls on the table pulls the balls behind it, so its tension is their mass times $a$.

Step 1. Situation at the instant. Ball 1 went over first. When the 6th ball has just left the table, balls $1$–$6$ hang and balls $7$–$10$ are on the table.

Step 2. Acceleration of the system. Driving force $=$ weight of $6$ balls; total mass $=10$ balls: $$a=\frac{6\times2\times10}{10\times2}=\frac{120}{20}=6\ \text{m s}^{-2}$$

Step 3. Isolate balls 8, 9, 10. They lie on the smooth table, and the only horizontal force on them is the tension in the string between balls 7 and 8: $$T=(3\times2)\times a=6\times6=36\ \text{N}$$

Answer. $36$ N. Count carefully which balls lie beyond the link: the string between 7 and 8 drags only 8, 9 and 10.

Concept. The $8$ kg box hangs from a movable pulley held by two segments of the string, so it is supported by $2T$ and moves half as far as the free end. The free end carries the $2$ kg box. Write the constraint first, then Newton's law for each box.

Step 1. Constraint. From the figure the string runs from the ceiling, under the movable pulley, over the fixed pulley, to the $2$ kg box. If the $8$ kg box goes down by $x$, both segments on the movable pulley lengthen by $x$, so the $2$ kg box rises by $2x$. Hence if the $8$ kg box has acceleration $a$ down, the $2$ kg box has $2a$ up.

Step 2. Isolate the $8$ kg box (with its pulley). Forces: $80$ N down, $2T$ up. Down positive: $$80-2T=8a\qquad(1)$$

Step 3. Isolate the $2$ kg box. Forces: $T$ up, $20$ N down. Up positive: $$T-20=2(2a)=4a\qquad(2)$$

Step 4. Solve. Multiply (2) by 2 and add to (1): $$80-40=8a+8a\ \Rightarrow\ a=2.5\ \text{m s}^{-2},\qquad T=30\ \text{N}$$

Step 5. Time to fall $20$ cm from rest. $$0.2=\tfrac12(2.5)t^2\ \Rightarrow\ t^2=0.16\ \Rightarrow\ t=0.4\ \text{s}$$

Answer. $0.4$ s, option D. Forgetting the factor 2 in the constraint is the usual slip.

Concept. The $2$ kg block on the smooth incline is pulled by the string; the other end goes over a fixed pulley and down to a movable pulley carrying the $4$ kg block, with the string's far end fixed to the ceiling. So the $4$ kg block is held by $2T$ and moves half as fast as the $2$ kg block.

Step 1. Constraint. If the $2$ kg block moves up the incline with acceleration $a$, the string feeds that length onto the two segments of the movable pulley, so the $4$ kg block descends with $a/2$.

Step 2. Isolate the $2$ kg block. Forces along the incline: $T$ up the incline, $2g\sin30^\circ=g$ down the incline ($N$ is perpendicular, and there is no friction). $$T-2g\sin30^\circ=2a\ \Rightarrow\ T=g+2a\qquad(1)$$

Step 3. Isolate the $4$ kg block with its pulley. Forces: $4g$ down, $2T$ up. Down positive: $$4g-2T=4\cdot\frac a2=2a\qquad(2)$$

Step 4. Solve. Put (1) into (2): $$4g-2g-4a=2a\ \Rightarrow\ 2g=6a\ \Rightarrow\ a=\frac g3$$

Answer. $\dfrac g3$, option B. This is the acceleration of the $2$ kg block; the $4$ kg block has $\dfrac g6$.

Concept. For light strings over light pulleys, string-length constraints link the accelerations. Write one constraint for each string, measuring lengths from the pulley it passes over, then eliminate the pulleys' own accelerations.

Step 1. Set up. Take downward accelerations as positive. From the figure: string 1 goes over the fixed pulley with $m_1$ on one side and pulley 2 on the other; string 2 goes over pulley 2 with $m_2$ on one side and pulley 3 on the other; string 3 goes over pulley 3 with $m_3$ and $m_4$. Let $b_2$, $b_3$ be the accelerations of pulleys 2 and 3.

Step 2. String over the fixed pulley. Its two hanging lengths add to a constant: $$a_1+b_2=0\qquad(1)$$

Step 3. String over pulley 2. Lengths are measured from pulley 2, so use accelerations relative to it: $$(a_2-b_2)+(b_3-b_2)=0\ \Rightarrow\ a_2+b_3=2b_2\qquad(2)$$

Step 4. String over pulley 3. $$(a_3-b_3)+(a_4-b_3)=0\ \Rightarrow\ b_3=\frac{a_3+a_4}{2}\qquad(3)$$

Step 5. Eliminate. From (1), $b_2=-a_1$. Put this and (3) into (2): $$a_2+\frac{a_3+a_4}{2}=-2a_1\ \Rightarrow\ 4a_1+2a_2+a_3+a_4=0$$

Answer. Option A. Shortcut (tension method): if string 3 has tension $T$, string 2 has $2T$ and string 1 has $4T$; so the masses feel $4T,2T,T,T$ and $\sum T_ia_i=0$ gives the coefficients $4,2,1,1$.

Concept. Kinetic friction is $\mu N$, and $N$ depends on the vertical part of the applied force. Pushing at an angle below the horizontal presses the block into the floor (larger $N$, more friction); pulling at an angle above lifts it (smaller $N$, less friction). The horizontal part of $F$ is the same in both cases.

Step 1. Components of $F$. $$F\cos30^\circ=20\times0.866=17.32\ \text{N},\qquad F\sin30^\circ=10\ \text{N}$$ Weight $mg=50$ N.

Step 2. Case (A), pushed. Vertically: $N_A=mg+F\sin30^\circ=60$ N, so friction $=0.2\times60=12$ N. $$a_A=\frac{17.32-12}{5}=1.06\ \text{m s}^{-2}$$

Step 3. Case (B), pulled. Vertically: $N_B=mg-F\sin30^\circ=40$ N, so friction $=0.2\times40=8$ N. $$a_B=\frac{17.32-8}{5}=1.86\ \text{m s}^{-2}$$

Step 4. Difference. $$a_B-a_A=\frac{12-8}{5}=0.8\ \text{m s}^{-2}$$

Answer. $0.8$ m s$^{-2}$, option C. Quick route: only the friction differs, by $\mu\cdot2F\sin30^\circ=0.2\times20=4$ N, giving $4/5=0.8$.

Concept. Static friction can act either way along the incline, up to a maximum $f_L=\mu mg\cos\theta$. At the largest push down the incline friction is at its limit and acts up; at the largest push up it is at its limit and acts down. Two equations, two unknowns ($mg$ and $\mu$).

Step 1. Largest force down the incline, $2$ N. The block tends to slide down, so friction acts up the incline at its limit: $$2+mg\sin30^\circ=f_L\qquad(1)$$

Step 2. Largest force up the incline, $10$ N. The block tends to slide up, so friction acts down: $$10=mg\sin30^\circ+f_L\qquad(2)$$

Step 3. Solve. Subtract (1) from (2): $$8=2mg\sin30^\circ=mg\ \Rightarrow\ mg=8\ \text{N}$$ Add (1) and (2): $12=2f_L$, so $f_L=6$ N.

Step 4. Coefficient of friction. $$\mu=\frac{f_L}{mg\cos30^\circ}=\frac{6}{8\times\frac{\sqrt3}{2}}=\frac{6}{4\sqrt3}=\frac{\sqrt3}{2}$$

Answer. $\dfrac{\sqrt3}{2}$, option D. Pattern: $F_{\text{up}}+F_{\text{down}}=2f_L$ and $F_{\text{up}}-F_{\text{down}}=2mg\sin\theta$.

Concept. On an incline of the same length, starting from rest, $s=\tfrac12at^2$ gives $t\propto\dfrac1{\sqrt a}$. Friction reduces the acceleration down the slope from $g\sin\theta$ to $g(\sin\theta-\mu\cos\theta)$, so comparing times compares accelerations.

Step 1. The two accelerations. Isolate the object; forces along the slope are $mg\sin\theta$ down and (on the rough plane) $\mu mg\cos\theta$ up. $$a_{\text{smooth}}=g\sin45^\circ,\qquad a_{\text{rough}}=g\sin45^\circ-\mu g\cos45^\circ$$

Step 2. Relate times. $t_{\text{rough}}=n\,t_{\text{smooth}}$ and $t\propto a^{-1/2}$, so $$\frac{a_{\text{smooth}}}{a_{\text{rough}}}=n^2$$

Step 3. Substitute. $\sin45^\circ=\cos45^\circ$ cancels: $$\frac{g\sin45^\circ}{g\sin45^\circ(1-\mu)}=n^2\ \Rightarrow\ \frac1{1-\mu}=n^2\ \Rightarrow\ \mu=1-\frac1{n^2}$$

Answer. Option D. For a general angle the result is $\mu=\tan\theta\left(1-\dfrac1{n^2}\right)$.

Concept. The block goes up and comes back over the same distance $d$, but friction always opposes its motion: it adds to $g\sin\theta$ going up and subtracts coming down. Use $v^2=2as$ for each leg and compare.

Step 1. Accelerations. Isolate the block; along the incline: $$\text{up (retardation): } a_1=g(\sin\theta+\mu\cos\theta),\qquad\text{down: } a_2=g(\sin\theta-\mu\cos\theta)$$

Step 2. Each leg over the same $d$. Going up, $v_0$ falls to zero; coming down, the block starts from rest and reaches $v_0/2$: $$v_0^2=2a_1d,\qquad \frac{v_0^2}{4}=2a_2d\ \Rightarrow\ a_1=4a_2$$

Step 3. Solve for $\mu$. $$\sin\theta+\mu\cos\theta=4\sin\theta-4\mu\cos\theta\ \Rightarrow\ 5\mu\cos\theta=3\sin\theta\ \Rightarrow\ \mu=\frac35\tan30^\circ$$ $$\mu=\frac{3}{5\sqrt3}=\frac{\sqrt3}{5}\approx0.346$$

Answer. $\mu\approx\dfrac{346}{1000}$, so $I=346$. General result: returning with $v_0/n$ gives $\mu=\tan\theta\,\dfrac{n^2-1}{n^2+1}$; here $n=2$ gives $\tfrac35\tan\theta$.

Concept. The hanging part pulls the chain over the edge with its weight; friction on the part lying on the table resists. The chain stays at rest as long as the pull does not exceed the limiting friction, so the longest hanging part makes them equal.

Step 1. Set up. Let the mass per unit length be $\lambda$, the total length $L=6$ m and the hanging length $x$. The part on the table has length $L-x$ and normal force $\lambda(L-x)g$.

Step 2. Limiting condition. $$\lambda xg=\mu\lambda(L-x)g\ \Rightarrow\ x=\mu(L-x)$$

Step 3. Solve. $$x(1+\mu)=\mu L\ \Rightarrow\ x=\frac{\mu L}{1+\mu}=\frac{0.5\times6}{1.5}=2\ \text{m}$$

Answer. $2$ m. Remember: the fraction hanging is $\dfrac{\mu}{1+\mu}$, not $\mu$.

Concept. On a curved surface, each point acts like a small incline whose slope is $\tan\theta=\dfrac{dy}{dx}$. A block stays at rest where $\tan\theta\le\mu$ (the same condition as the angle of repose). The curve gets steeper with height, so the highest resting point is where $\dfrac{dy}{dx}=\mu$.

Step 1. Slope of the curve. $$y=\frac{x^2}{4}\ \Rightarrow\ \frac{dy}{dx}=\frac x2$$

Step 2. Limiting point. Isolate the block: along the surface, $mg\sin\theta$ must not exceed $\mu mg\cos\theta$, i.e. $\tan\theta\le\mu$. $$\frac x2\le0.5\ \Rightarrow\ x\le1\ \text{m}$$

Step 3. Height there. $$y_{\max}=\frac{1^2}{4}=0.25\ \text{m}=25\ \text{cm}$$

Answer. $25$ cm. The question asks for height ($y$), not horizontal distance.

Concept. The bag is dropped with no horizontal speed onto a belt moving at $v$. Kinetic friction from the belt acts forward on the bag until the bag reaches the belt's speed. The distance asked is the slip relative to the belt, which is easiest in the belt's frame (inertial, as the belt moves uniformly).

Step 1. Friction and acceleration. Isolate the bag: normal force $mg$, so friction $\mu mg$ and $$a=\mu g=0.4\times10=4\ \text{m s}^{-2}$$

Step 2. In the belt's frame. The bag starts moving backward at $2$ m s$^{-1}$, and friction slows it at $4$ m s$^{-2}$ until it is at rest on the belt.

Step 3. Slip distance. $$s=\frac{u^2}{2a}=\frac{2^2}{2\times4}=0.5\ \text{m}$$

Answer. $0.5$ m, option B. In the ground frame the bag moves $0.5$ m forward while the belt moves $1$ m in the same time ($t=0.5$ s); the difference, $0.5$ m, is the slip.

Concept. The force is applied to the lower block $M$. The upper block $m$ is dragged along only by static friction from $M$, and this friction cannot exceed $\mu mg$. So $m$ (and hence the whole system, if they move together) can accelerate at most at $\mu g$.

Step 1. Isolate the upper block $m$. Horizontal force: friction $f$ forward. Vertical: $N=mg$. $$f=ma\le\mu mg\ \Rightarrow\ a\le\mu g=0.5\times9.8=4.9\ \text{m s}^{-2}$$

Step 2. Whole system. Moving together on a smooth table, the only external horizontal force is $F$: $$F=(m+M)a$$

Step 3. Largest $F$. $$F_{\max}=(2+8)\times4.9=49\ \text{N}$$

Answer. $49$ N, option C. Using $\mu mg=9.8$ N as the answer is a trap: that is the friction on $m$, not the applied force.

Concept. As in the smooth-floor case, $A$ is accelerated only by friction from $B$, so it can have at most $\mu_1 g$. Now the floor is rough, so $B$ also feels kinetic friction from the table, which the force $F$ must overcome.

Step 1. Isolate $A$. The only horizontal force is friction from $B$: $$m_Aa\le\mu_1m_Ag\ \Rightarrow\ a_{\max}=\mu_1g=0.2\times10=2\ \text{m s}^{-2}$$

Step 2. Friction from the table on $B$. The table supports both blocks, so its normal force is $(m_A+m_B)g=40$ N: $$f_2=\mu_2(m_A+m_B)g=0.2\times40=8\ \text{N}$$

Step 3. Both blocks together. External horizontal forces: $F$ forward, $f_2$ backward: $$F-f_2=(m_A+m_B)a_{\max}\ \Rightarrow\ F-8=4\times2\ \Rightarrow\ F=16\ \text{N}$$

Answer. $16$ N, option A. Shortcut: $F_{\max}=(\mu_1+\mu_2)(m_A+m_B)g=0.4\times40$.

Concept. $M$ moves down its face and $m$ moves up its face, joined by one string over an ideal pulley (same tension, same acceleration $a$). Kinetic friction on each block acts opposite to its own motion. Isolate each block, resolve along its face, and apply Newton's law.

Step 1. Trigonometric values. $\sin53^\circ=\cos37^\circ=0.8$, $\cos53^\circ=\sin37^\circ=0.6$.

Step 2. Isolate $M=10$ kg (on the $53^\circ$ face, moving down). Along the face: $Mg\sin53^\circ=80$ N down, friction $\mu Mg\cos53^\circ=0.25\times60=15$ N up, tension $T$ up. $$80-15-T=10\times2\ \Rightarrow\ T=45\ \text{N}$$

Step 3. Isolate $m$ (on the $37^\circ$ face, moving up). Along the face: $T$ up; $mg\sin37^\circ=6m$ down; friction $\mu mg\cos37^\circ=0.25\times8m=2m$ down. $$T-6m-2m=m\times2\ \Rightarrow\ 45=10m$$

Step 4. Solve. $$m=4.5\ \text{kg}$$

Answer. $4.5$ kg, option B. Always fix the direction of motion first; it decides which way each friction force points.

Concept. A lift moving with uniform velocity has no acceleration, so it is an inertial frame: everything inside behaves exactly as on the ground. On a smooth incline the block slides down with $a=g\sin\theta$, and from rest $l=\tfrac12at^2$.

Step 1. Time in terms of angle. Isolate the block: along the smooth incline only $mg\sin\theta$ acts, so $a=g\sin\theta$ and $$l=\tfrac12g\sin\theta\,t^2\ \Rightarrow\ t=\sqrt{\frac{2l}{g\sin\theta}}\propto\frac1{\sqrt{\sin\theta}}$$

Step 2. Ratio for the two angles. The length $l$ is the same: $$\frac{t_2}{t_1}=\sqrt{\frac{\sin30^\circ}{\sin45^\circ}}=\sqrt{\frac{0.5}{0.707}}=\sqrt{0.707}\approx0.841$$

Step 3. New time. $$t_2=2\times0.841\approx1.68\ \text{s}$$

Answer. $1.68$ s, option C. The speed $v$ of the lift plays no part; only acceleration of the frame would.

Concept. In the car's frame, which accelerates at $\vec a$, the bob feels a pseudo force $-m\vec a$ besides gravity. The bob hangs at rest along the effective gravity $\vec g_{\text{eff}}=\vec g-\vec a$, so the string's direction is the direction of this vector.

Step 1. Components. Take $x$ horizontal (in the direction the car moves) and $y$ vertically up. The car's acceleration is $10$ m s$^{-2}$ up the $30^\circ$ incline: $$\vec a=(10\cos30^\circ,\ 10\sin30^\circ)=(5\sqrt3,\ 5),\qquad \vec g=(0,\ -10)$$

Step 2. Effective gravity. $$\vec g_{\text{eff}}=\vec g-\vec a=(-5\sqrt3,\ -15)$$ It points down and backward.

Step 3. Angle with the vertical. $$\tan\theta=\frac{|g_{\text{eff},x}|}{|g_{\text{eff},y}|}=\frac{5\sqrt3}{15}=\frac{1}{\sqrt3}\ \Rightarrow\ \theta=30^\circ$$

Answer. $30^\circ$. Using $\tan\theta=a/g=1$ ($45^\circ$) is wrong here because $\vec a$ is not horizontal; its vertical part increases the effective gravity.

Concept. Work in the frame of the incline. It accelerates to the left at $a_0$, so the block feels a pseudo force $ma_0$ to the right. In this frame the block slides along the incline from rest, so find its acceleration along the incline and use $s=\tfrac12at^2$ with the incline's length.

Step 1. Geometry. From the figure the incline rises to the right, the block starts at the top (right end), and the base is $L$. So the length along the slope is $$\ell=\frac{L}{\cos\theta}$$ "Down the incline" points to the lower left.

Step 2. Forces along the incline (down the slope positive).

• Gravity: $mg\sin\theta$ down the slope.
• Pseudo force $ma_0$ to the right: its component along the slope is $ma_0\cos\theta$, pointing up the slope.
• Normal force: no component along the slope.

$$a_{\text{rel}}=g\sin\theta-a_0\cos\theta$$

Step 3. Time from rest. $$\frac{L}{\cos\theta}=\tfrac12(g\sin\theta-a_0\cos\theta)t^2\ \Rightarrow\ t^2=\frac{2L}{g\sin\theta\cos\theta-a_0\cos^2\theta}$$

Step 4. Match the options. Multiply top and bottom by 2 and use $2\sin\theta\cos\theta=\sin2\theta$, $2\cos^2\theta=1+\cos2\theta$: $$t=\sqrt{\frac{4L}{g\sin2\theta-a_0(1+\cos2\theta)}}$$

Answer. Option B. Option A uses $L$ as if it were the slope length.

Concept. Work in the cube's frame. It accelerates to the right at $g/2$, so each block feels a pseudo force $m\dfrac g2$ to the left. In this frame both blocks are at rest, so the forces on each balance. On $Q$ (on the right side face) the pseudo force presses it into the face and so produces the normal force, which limits friction.

Step 1. Pseudo forces. With $g=10$ m s$^{-2}$, $g/2=5$ m s$^{-2}$: $$P:\ 2\times5=10\ \text{N (left)},\qquad Q:\ 1.5\times5=7.5\ \text{N (left)}$$

Step 2. Isolate $P$ (smooth top face). Horizontal forces: tension $T$ to the right (towards the pulley) and the pseudo force to the left. $$T=10\ \text{N}$$

Step 3. Isolate $Q$, horizontally. The pseudo force ($7.5$ N, left) pushes $Q$ against the face; the face pushes back: $$N=7.5\ \text{N}$$

Step 4. Isolate $Q$, vertically. Weight $15$ N down, tension $10$ N up. The rest must come from friction, acting up: $$f=15-10=5\ \text{N}$$

Step 5. Condition for no slipping. $$f\le\mu N\ \Rightarrow\ 5\le7.5\mu\ \Rightarrow\ \mu\ge\frac23\approx0.67$$

Answer. $0.67$, option B (the least value that keeps the blocks stationary).

Concept. Solve the relative motion in $M$'s frame, where each small block gets a pseudo force. Then return to the ground frame: the only external horizontal force on the whole system is $F$, so $F=\sum m_ia_{i,x}$ using ground-frame accelerations.

Step 1. Constraint and directions. Let $M$ accelerate to the right at $a$. From the figure, $m_1$ is on top of $M$ joined over the pulley at $M$'s right edge to $m_2$, which hangs against $M$'s right face and so moves horizontally with $M$. If $m_2$ rises at $2$ m s$^{-2}$ relative to $M$, then $m_1$ slides to the left (away from the pulley) at $2$ m s$^{-2}$ relative to $M$.

Step 2. Isolate $m_2$, vertically. Forces: $T$ up, $m_2g$ down; acceleration $2$ m s$^{-2}$ up (vertical motion is the same in both frames): $$T-m_2g=m_2(2)\ \Rightarrow\ T=20(10+2)=240\ \text{N}$$

Step 3. Isolate $m_1$ in $M$'s frame. Horizontal forces: pseudo force $m_1a$ to the left, tension $T$ to the right. It accelerates left at $2$ m s$^{-2}$ relative to $M$: $$m_1a-T=m_1(2)\ \Rightarrow\ 10a-240=20\ \Rightarrow\ a=26\ \text{m s}^{-2}$$

Step 4. Ground-frame horizontal accelerations. $$M:\ 26,\qquad m_2:\ 26,\qquad m_1:\ 26-2=24\ \text{m s}^{-2}$$

Step 5. Whole system. The floor is smooth; tension and contact forces between the bodies are internal. So $$F=100(26)+20(26)+10(24)=2600+520+240=3360\ \text{N}$$

Answer. $3360$ N, option A. The usual error is to give $m_1$ the same $26$ m s$^{-2}$, which gives $3380$ N (option B).

Concept. The coin moves in a circle with the table, and the centripetal force $m\omega^2r$ is supplied by static friction, which can be at most $\mu mg$. The coin just slips when the needed force equals the maximum friction.

Step 1. Slipping condition. Isolate the coin: $N=mg$ vertically, friction towards the centre horizontally. $$m\omega^2r=\mu mg\ \Rightarrow\ r=\frac{\mu g}{\omega^2}\propto\frac1{\omega^2}$$

Step 2. Halve $\omega$. $\mu$ and $g$ do not change, so $$\frac{r_2}{r_1}=\left(\frac{\omega_1}{\omega_2}\right)^2=2^2=4\ \Rightarrow\ r_2=4\times1=4\ \text{cm}$$

Answer. $4$ cm, option B. Halving $\omega$ cuts the needed force by four, so the coin can sit four times farther out.

Concept. A body pressed against the inside wall of a rotating drum moves in a horizontal circle. The wall's normal force points to the axis and supplies the centripetal force. Friction from the wall acts upward and must hold the weight, and it can be at most $\mu N$.

Step 1. Horizontal (towards the axis). $$N=m\omega^2r$$

Step 2. Vertical. For the mass not to slip down: $$mg\le\mu N=\mu m\omega^2r$$

Step 3. At the minimum speed the inequality is an equality. $$\mu=\frac{g}{\omega^2r}=\frac{10}{5^2\times4}=\frac{10}{100}=0.1$$

Answer. $0.1$, option A. The mass ($0.5$ kg) cancels and is not needed.

Concept. The outer rail is raised so that the normal force from the track, tilted inward, supplies the centripetal force by itself (no sideways push on the wheel flanges). This gives $\tan\theta=\dfrac{v^2}{rg}$. For a small angle, the height difference between rails is $h=l\sin\theta\approx l\tan\theta$.

Step 1. Banking angle. Isolate the train: $N\cos\theta=mg$ vertically and $N\sin\theta=\dfrac{mv^2}{r}$ horizontally. Dividing, $$\tan\theta=\frac{v^2}{rg}=\frac{12^2}{400\times10}=\frac{144}{4000}=0.036$$

Step 2. Height of the outer rail. The rails are $l=1.5$ m apart along the tilted track, and $\theta$ is small: $$h=l\sin\theta\approx l\tan\theta=1.5\times0.036=0.054\ \text{m}$$

Answer. $5.4$ cm, option B.

Concept. At the maximum safe speed the car tends to slide outward, so static friction acts down the slope at its limit $\mu N$. The normal force and friction together must give zero vertical force and the centripetal force horizontally.

Step 1. Forces on the car. Weight $mg$ down; normal $N$ perpendicular to the road; friction $\mu N$ down the slope.

Step 2. Vertical balance. $$N\cos\theta-\mu N\sin\theta=mg$$

Step 3. Horizontal (towards the centre). $$N\sin\theta+\mu N\cos\theta=\frac{mv^2}{r}$$

Step 4. Divide to remove $N$ and $m$. $$\frac{v^2}{rg}=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}=\frac{\tan\theta+\mu}{1-\mu\tan\theta}$$

Step 5. Substitute. $\tan30^\circ=\dfrac{1}{1.73}=0.578$: $$v^2=300\times10\times\frac{0.578+0.2}{1-0.2\times0.578}=3000\times\frac{0.778}{0.884}\approx2639$$ $$v\approx51.4\ \text{m s}^{-1}$$

Answer. $51.4$ m/s, option B. The car's mass ($800$ kg) cancels. For $v_{\min}$, friction acts up the slope: change the signs of $\mu$.

Concept. This is a conical pendulum: the mass moves in a horizontal circle while the string sweeps a cone. The horizontal part of the tension supplies the centripetal force. The circle's radius is $L\sin\theta$, not $L$.

Step 1. Forces on $M$. Tension $T$ along the string (at $\theta$ to the vertical) and weight $Mg$. Radius of the circle $R=L\sin\theta$.

Step 2. Horizontal (towards the axis). $$T\sin\theta=M\omega^2R=M\omega^2L\sin\theta\ \Rightarrow\ T=M\omega^2L$$ The angle cancels, so we do not need it.

Step 3. Angular speed. $f=\dfrac3\pi$ rotations per second: $$\omega=2\pi f=2\pi\times\frac3\pi=6\ \text{rad s}^{-1}$$

Step 4. Tension. $$T=M(6)^2L=36\,ML$$

Answer. $36$. The vertical equation $T\cos\theta=Mg$ would give $\theta$, but it is not needed here.

Concept. In the car's frame (rotating with it), the bob is at rest under its weight $mg$ downward, the centrifugal force $\dfrac{mv^2}{r}$ outward, and the tension. The string lines up with the resultant of weight and centrifugal force.

Step 1. Convert the speed. $$v=54\ \text{km h}^{-1}=54\times\frac{5}{18}=15\ \text{m s}^{-1}$$

Step 2. Balance on the bob. With the string at angle $\theta$ to the vertical: $$T\cos\theta=mg,\qquad T\sin\theta=\frac{mv^2}{r}$$

Step 3. Divide. $$\tan\theta=\frac{v^2}{rg}=\frac{15^2}{20\times10}=\frac{225}{200}=1.125$$

Answer. $\theta=\tan^{-1}(1.125)$, option C. The bob leans outward, away from the centre of the turn.

Concept. The object moves in a circle on a smooth table, and the stretched spring supplies the centripetal force. Stretching the spring by $x$ makes the radius $l_0+x$, so the radius itself depends on the answer.

Step 1. Forces on the object. Horizontally, only the spring force $kx$, towards the fixed end (the centre). Vertically, weight and normal force cancel.

Step 2. Newton's second law towards the centre. $$kx=m\omega^2(l_0+x)$$

Step 3. Solve for $x$. $$kx-m\omega^2x=m\omega^2l_0\ \Rightarrow\ x=\frac{m\omega^2l_0}{k-m\omega^2}$$

Answer. Option C. Using $l_0$ as the radius gives $m\omega^2l_0/k$, which is only the small-$\omega$ approximation. If $m\omega^2\ge k$ the spring cannot hold the object.

Concept. The bead moves in a horizontal circle about the vertical axis AB. The smooth wire can only push along the radius of the ring, so the normal force points from P towards the centre O. Its vertical part balances gravity and its horizontal part supplies the centripetal force.

Step 1. Geometry. From the figure, P is below O at a horizontal distance $\dfrac r2$ from the axis. Let OP make angle $\theta$ with the downward vertical. Then the radius of the bead's circle is $$\rho=r\sin\theta=\frac r2\ \Rightarrow\ \sin\theta=\frac12,\ \theta=30^\circ$$

Step 2. Forces on the bead. Weight $mg$ down; normal $N$ along PO, which points up and towards the axis.

Step 3. Vertical balance. $$N\cos\theta=mg$$

Step 4. Horizontal (towards the axis). $$N\sin\theta=m\omega^2(r\sin\theta)\ \Rightarrow\ N=m\omega^2r$$

Step 5. Combine. $$m\omega^2r\cos\theta=mg\ \Rightarrow\ \omega^2=\frac{g}{r\cos30^\circ}=\frac{2g}{\sqrt3\,r}$$

Answer. Option C. The bead can sit off the bottom only if $\omega^2>g/r$; here $\omega^2=1.15\,g/r$, which satisfies this.

Concept. In a vertical circle the tension is greatest at the bottom and least at the top. At each point, the net force towards the centre equals $mv^2/l$. Energy conservation links the speeds at the top and bottom.

Step 1. Speeds. Let $v$ be the speed at the top. The bob falls $2l$ to the bottom: $$u^2=v^2+2g(2l)=v^2+4gl$$

Step 2. Tension at the top. Both $T$ and $mg$ point towards the centre (down): $$T_{\min}+mg=\frac{mv^2}{l}\ \Rightarrow\ T_{\min}=\frac{mv^2}{l}-mg$$

Step 3. Tension at the bottom. $T$ points to the centre (up), $mg$ away: $$T_{\max}-mg=\frac{mu^2}{l}\ \Rightarrow\ T_{\max}=\frac{m(v^2+4gl)}{l}+mg=\frac{mv^2}{l}+5mg$$

Step 4. Use the ratio. With $l=1$ m and $g=10$ m s$^{-2}$ (the $m$ cancels): $$\frac{v^2+50}{v^2-10}=5\ \Rightarrow\ v^2+50=5v^2-50\ \Rightarrow\ 4v^2=100\ \Rightarrow\ v^2=25$$

Answer. $v=5$ m s$^{-1}$. Check: $v^2=25>gl=10$, so the string stays taut at the top. Also $T_{\max}-T_{\min}=6mg$ always.