Question

a block of mass ( m ) slides with an initial velocity ( v_0 ) along a rough surface where the coefficient of kinetic friction between the block and the surface is ( mu ). the box comes to rest after sliding a distance ( d_0 ). a new block of unknown mass slides with an initial velocity of ( 2v_0 ) across a surface where the coefficient of kinetic friction between the new block and the surface is ( \frac{mu}{2} ). which of the following expressions represents the distance the new block slides before coming to rest in terms of ( d_0 )?

Explanation:

Step1: Apply work - energy theorem for the first block

The work - energy theorem states that \(W=\Delta K\). The initial kinetic energy \(K_{i}=\frac{1}{2}mv_{0}^{2}\), and the final kinetic energy \(K_{f} = 0\). The work done by friction \(W=-F_{f}d_{0}\), where the frictional force \(F_{f}=\mu mg\). So, \(-\mu mgd_{0}=0 - \frac{1}{2}mv_{0}^{2}\), which simplifies to \(\mu gd_{0}=\frac{1}{2}v_{0}^{2}\).

Step2: Apply work - energy theorem for the second block

Let the mass of the second block be \(M\), initial velocity \(v = 2v_{0}\), and coefficient of kinetic friction \(\mu'=\frac{\mu}{2}\). The initial kinetic energy \(K_{i}=\frac{1}{2}M(2v_{0})^{2}=2Mv_{0}^{2}\), and the final kinetic energy \(K_{f} = 0\). The work done by friction \(W=-F_{f}'d\), where \(F_{f}'=\mu'Mg=\frac{\mu}{2}Mg\). So, \(-\frac{\mu}{2}Mgd=0 - 2Mv_{0}^{2}\), which simplifies to \(\frac{\mu}{2}gd = 2v_{0}^{2}\).

Step3: Relate the two equations

From \(\mu gd_{0}=\frac{1}{2}v_{0}^{2}\), we have \(v_{0}^{2}=2\mu gd_{0}\). Substitute \(v_{0}^{2}\) into \(\frac{\mu}{2}gd = 2v_{0}^{2}\):

$$
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Answer:

D. \(8d_{0}\)