Question #100851
A wooden block slides directly down an inclined plane with a constant speed of 6.0 m/s. What is the coefficient of kinetic friction if the incline is angled at 25º.
1
Expert's answer
2020-01-02T11:11:48-0500

The free body diagram of the considered problem is (note: figure is a courtesy of https://sciencenotes.org/friction-example-problem-sliding-inclined-plane/):



As long as the velocity is constant, the net force should be equal to 0:

N+W+Ff=0\vec{N} + \vec{W} + \vec{F}_f = 0

Projecting this equation on the x- and y-axis, we obtain:

N=WcosθN=mgcos25Wsinθ=Frmgsin25=μNN = W \cos{\theta} \quad \Rightarrow N = mg \cos{25^\circ} \\ W \sin{\theta} = F_r \quad \Rightarrow mg \sin{25^\circ} = \mu N

Combining the expressions together, we derive:

sin25=μcos25μ=tan25\sin{25^\circ} = \mu \cos{25^\circ} \quad \Rightarrow \mu = \tan{25^\circ} \approx 0.47


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