A $1$ kg block subjected to two simultaneous forces $(2\hat{i} + 3\hat{j} + 4\hat{k})$ N and $(3\hat{i} - \hat{j} - 2\hat{k})$ N is moved a distance of $25$ m along $(3\hat{i} - 4\hat{j})$ direction. The work done in this process is ______ J.
Numerical answer type
Answer and solution
Answer: 35
Concept. Work done by a constant force is the dot product $W=\vec F\cdot\vec s$. Two forces act at once, so we first add them to get the net force, and the displacement must be written as a vector before the dot product can be taken.
Step 1. Net force. Add the two forces component by component: $$\vec F=(2+3)\hat i+(3-1)\hat j+(4-2)\hat k=5\hat i+2\hat j+2\hat k\ \text{N}$$
Step 2. Displacement as a vector. The block moves $25$ m along $3\hat i-4\hat j$. The unit vector in that direction is $\dfrac{3\hat i-4\hat j}{\sqrt{3^2+4^2}}=\dfrac{3\hat i-4\hat j}{5}$, so $$\vec s=25\times\frac{3\hat i-4\hat j}{5}=15\hat i-20\hat j\ \text{m}$$
Step 3. Dot product. Multiply matching components and add. The displacement has no $\hat k$ part, so the $z$-component of the force does no work: $$W=(5)(15)+(2)(-20)+(2)(0)=75-40=35\ \text{J}$$
Answer. $35$ J. A common slip is to dot with $3\hat i-4\hat j$ itself (giving $7$ J) instead of the actual $25$ m displacement vector.
Point to note. Add the two forces, turn the displacement into a vector ($25\times$ unit vector), then dot.