A rope tied to a body is pulled, causing the body to accelerate. But according to Newton’s third law, the body pulls back on the rope with a force of equal magnitude and opposite direction. Is the total work done then zero? If so, how can the body’s kinetic energy change? Explain.

The work energy theorem tells that the work done on the body results in a change in its kinetic energy.

$$\begin{gathered} {\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{∆}\mathbf{K} \end{gathered}$$

The particle will undergo a displacement when the force acts on a particle, and then the particle's kinetic energy changes by an amount, which is equal to the total work done on the particle by all the forces.

Here, the total work done on the body is not zero, because if the rope pulls the body by applying force on the body and according to Newton's third law the body in turn pulls the rope at the point of contact of the rope with the body.

So, the net force at the point of contact is zero. But there will be net force acting on the system. Therefore, the required change in kinetic energy is equal to the total work done on the system is as follows.

$$\begin{gathered} ∆KE=W \\ =\frac{1}{2}m{v}^2 \end{gathered}$$

Therefore, the kinetic energy is $\frac{1}{2}m{v}^2$.