Question

A force F = − k ( x ˆ i + y ˆ j ) [where k is a positive constant] acts on a particle moving in the x-y plane.Starting from origin, the particle is taken to (a, a) and then to ( a 2 , 0. Find the total work done

Answer 1

Answer on Question#56926 - Physics - Mechanics - Relativity

A force $F = -k(x\boldsymbol{i} + y\boldsymbol{j})$ [where $k$ is a positive constant] acts on a particle moving in the x-y plane. Starting from origin, the particle is taken to $(a, a)$ and then to $(a^2, 0)$. Find the total work done.

Solution:

Force $F$ is potential force and its potential is given by

$$U = \frac{k(x^2 + y^2)}{2}$$

Indeed

$$F = -\nabla U = -\frac{\partial}{\partial x} \left( \frac{k(x^2 + y^2)}{2} \right) \boldsymbol{i} - \frac{\partial}{\partial y} \left( \frac{k(x^2 + y^2)}{2} \right) \boldsymbol{j} = -k(x\boldsymbol{i} + y\boldsymbol{j})$$

Thus the total work $W$ done is given by the difference of potential energies between initial (origin) and final $((a^2, 0))$ points:

$$W = U(a^2, 0) - U(0, 0) = \frac{k((a^2)^2 + 0^2)}{2} - 0 = \frac{k a^4}{2}$$

Answer: $\frac{k a^4}{2}$.