Question

1. a box with mass of 75.0 kg slides down on an incline plane with an angle of 15° and at the bottom of incline plane its speed is precisely 90% of what it would have been if the slide had been frictionless. calculate the coefficient of kinetic friction between the slide and the box.

Explanation:

Step1: Find the acceleration without friction

Using Newton's second law \(F = ma\). On a frictionless incline, the net force \(F_{net}=mg\sin\theta\). So, \(a_{1}=g\sin\theta\).
Using the kinematic equation \(v^{2}=v_{0}^{2}+2aL\) (initial velocity \(v_{0} = 0\)), \(v_{1}^{2}=2a_{1}L=2gL\sin\theta\).

Step2: Find the acceleration with friction

Let the acceleration with friction be \(a_{2}\). The normal force \(N = mg\cos\theta\), and the frictional force \(f=\mu_{k}N=\mu_{k}mg\cos\theta\).
Using Newton's second law \(mg\sin\theta-\mu_{k}mg\cos\theta=ma_{2}\), so \(a_{2}=g(\sin\theta-\mu_{k}\cos\theta)\).
Using the kinematic equation \(v^{2}=v_{0}^{2}+2aL\) (initial velocity \(v_{0} = 0\)), \(v_{2}^{2}=2a_{2}L = 2gL(\sin\theta-\mu_{k}\cos\theta)\).

Step3: Use the given speed relation

Given \(v_{2}=0.9v_{1}\). Squaring both sides, \(v_{2}^{2}=0.81v_{1}^{2}\).
Substitute \(v_{1}^{2}=2gL\sin\theta\) and \(v_{2}^{2}=2gL(\sin\theta-\mu_{k}\cos\theta)\) into \(v_{2}^{2}=0.81v_{1}^{2}\):

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Substitute \(\theta = 15^{\circ}\), \(\tan15^{\circ}=2 - \sqrt{3}\approx0.268\).

$$ \mu_{k}=0.19\times0.268 = 0.051 $$

Answer:

\(0.051\)