A block of mass m moving with speed v compres | vedclass

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Question

(A) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the spring force.The initial kinetic energy is $K_i

Vedclass · Accepted Answer

$\frac{3mv^2}{4x^2}$

Vedclass · Answer

(A) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the spring force.The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.The final speed is $v_f = \frac{v}{2}$,so the final kinetic energy is $K_f = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{8}mv^2$.The work done by the spring force when it is compressed by distance $x$ is $W = -\frac{1}{2}kx^2$.Applying the theorem: $K_f - K_i = W$.$\frac{1}{8}mv^2 - \frac{1}{2}mv^2 = -\frac{1}{2}kx^2$.Multiplying by $8$: $mv^2 - 4mv^2 = -4kx^2$.$-3mv^2 = -4kx^2$.Solving for $k$: $k = \frac{3mv^2}{4x^2}$.

$A$ block of mass $m$ moving with speed $v$ compresses a spring through distance $x$ before its speed is halved. What is the value of the spring constant $k$?

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