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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) According to the work-energy theorem,the work done by the spring force is equal to the change in kinetic energy of the block.The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.The final kinetic energy when the speed is halved $(v' = v/2)$ is $K_f = \frac{1}{2}m(v/2)^2 = \frac{1}{8}mv^2$.The work done by the spring force is $W = -\Delta PE = -\frac{1}{2}kx^2$.By the work-energy theorem,$W = K_f - K_i$.$-\frac{1}{2}kx^2 = \frac{1}{8}mv^2 - \frac{1}{2}mv^2$.$-\frac{1}{2}kx^2 = \frac{mv^2 - 4mv^2}{8} = -\frac{3mv^2}{8}$.$\frac{1}{2}kx^2 = \frac{3mv^2}{8}$.$k = \frac{3mv^2}{4x^2}$.

$A$ 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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