Question

question 26
1 pts
a box has a mass of m and an initial velocity of \v\ m/s. it moves across a rough, flat surface and eventually comes to a stop after traveling a distance of \d\ meters. if the mass is doubled to \2m\ what will be the new distance it moves across the ground before stopping in terms of the original distance \d\?
4d
d
d/4
2d
d/2
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Explanation:

Step1: Calculate the initial kinetic energy

The initial kinetic energy \(K_{i}=\frac{1}{2}mv^{2}\).

Step2: Calculate the work done by friction

The work done by friction \(W = F_{f}d\), where \(F_{f}=\mu mg\) ( \(\mu\) is the coefficient of friction). So \(W=\mu mgd\).
When the box stops, \(K_{i} = W\), \(\frac{1}{2}mv^{2}=\mu mgd\), and we can get \(d=\frac{v^{2}}{2\mu g}\).

Step3: Analyze the case when mass is \(2m\)

For mass \(M = 2m\), the initial kinetic energy \(K_{i}'=\frac{1}{2}(2m)v^{2}=mv^{2}\).
The frictional force \(F_{f}'=\mu(2m)g\). Let the new distance be \(d'\). Then \(K_{i}'=F_{f}'d'\), \(mv^{2}=\mu(2m)gd'\).
Solve for \(d'\): \(d'=\frac{v^{2}}{2\mu g}\).

Answer:

\(d\)