A block of mass m moving with a velocity v_0 | vedclass

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Question

(B) For observer $A$ (ground frame),the block moves by a distance $x$ until it comes to rest.The force exerted by the spring on the block is $F = -kx$

Vedclass · Accepted Answer

$ - \frac{1}{2}mv_0^2$

Vedclass · Answer

(B) For observer $A$ (ground frame),the block moves by a distance $x$ until it comes to rest.The force exerted by the spring on the block is $F = -kx$ (in the opposite direction of motion).The work done by the spring force on the block is $W = \int_{0}^{x} (-kx) dx = -\frac{1}{2}kx^2$.According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{net} = \Delta K = K_f - K_i = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$.Since the surface is smooth and there is no other horizontal force,the work done by the normal reaction $N$ of the spring is the only work done on the block.Therefore,the work done by the normal reaction $N$ is $-\frac{1}{2}mv_0^2$.

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