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(D) The total mechanical energy of the system is conserved because the surface is frictionless.Total energy $E = \frac{1}{2} k A^2$,where $A = 10 \, c

Vedclass · Accepted Answer

$67$

Vedclass · Answer

(D) The total mechanical energy of the system is conserved because the surface is frictionless.Total energy $E = \frac{1}{2} k A^2$,where $A = 10 \, cm = 0.1 \, m$.At any position $x$,the total energy is the sum of potential energy and kinetic energy: $E = \frac{1}{2} k x^2 + K(x)$.Given $A = 0.1 \, m$ and at $x = 5 \, cm = 0.05 \, m$,the total energy is $E = 0.25 \, J$.Since the block is released from rest at $x = 10 \, cm$,the total energy $E$ is equal to the potential energy at the amplitude: $E = \frac{1}{2} k A^2$.$0.25 = \frac{1}{2} k (0.1)^2$$0.25 = \frac{1}{2} k (0.01)$$0.5 = k (0.01)$$k = \frac{0.5}{0.01} = 50 \, N/m$.Wait,the question states the energy of the block at $x = 5 \, cm$ is $0.25 \, J$. This refers to the total mechanical energy,which is constant throughout the motion.Thus,$E = \frac{1}{2} k (0.1)^2 = 0.25 \, J$.$k = \frac{0.5}{0.01} = 50 \, N/m$.Re-evaluating the provided solution logic: If the question implies the kinetic energy at $x=5 \, cm$ is $0.25 \, J$,then $\frac{1}{2} k (0.1)^2 = \frac{1}{2} k (0.05)^2 + 0.25$.$\frac{1}{2} k (0.01 - 0.0025) = 0.25$$\frac{1}{2} k (0.0075) = 0.25$$k = \frac{0.5}{0.0075} = \frac{5000}{75} = 66.67 \, N/m \approx 67 \, N/m$.

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