A spring-block system is placed on a rough ho | vedclass

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(C) Let the initial elongation be $x_0$. The block is released from rest. The spring force is $F_s = Kx$ and the maximum static friction is $f_{max} =

Vedclass · Accepted Answer

The block will come to rest when the work done by friction becomes equal to the change in energy stored in the spring.

Vedclass · Answer

(C) Let the initial elongation be $x_0$. The block is released from rest. The spring force is $F_s = Kx$ and the maximum static friction is $f_{max} = \mu mg$.For the block to move,the spring force must exceed the friction force. The block will stop when the spring force is less than or equal to the maximum static friction,i.e.,$|Kx| \le \mu mg$,or $|x| \le \frac{\mu mg}{K}$.By the Work-Energy Theorem,the work done by all forces equals the change in kinetic energy. Since the block starts and ends at rest,$\Delta KE = 0$. Thus,$W_{spring} + W_{friction} = 0$.$W_{spring} = -\Delta U = -(\frac{1}{2}Kx_f^2 - \frac{1}{2}Kx_0^2)$.$W_{friction} = -\int_{x_0}^{x_f} \mu mg dx = -\mu mg(x_0 - x_f)$ (assuming motion towards the mean position).Equating the work done: $\frac{1}{2}K(x_0^2 - x_f^2) = \mu mg(x_0 - x_f)$.Dividing by $(x_0 - x_f)$ (since $x_0 
eq x_f$): $\frac{1}{2}K(x_0 + x_f) = \mu mg$,which gives $x_f = \frac{2\mu mg}{K} - x_0$.Since $x_0 > \frac{\mu mg}{K}$,then $x_f Thus,the block does not cross the mean position,and the forces are not necessarily balanced at the stopping point. The work-energy relation is the fundamental condition for the stopping point.

$A$ spring-block system is placed on a rough horizontal floor. The block is pulled towards the right to give the spring an elongation $x_0$ such that $\frac{\mu mg}{K} < x_0 < \frac{2\mu mg}{K}$,and then released. Which of the following statements is correct?

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