Kevin S. Huang

2007: Problem 26

Topic: Dynamics
Concepts: 5 kinematics equations, Inclined plane, Newton’s laws

Solution:

Applying Newton’s 2nd law to the sled,

mg\sin\theta-f=ma

Since $f=\mu N=\mu mg\cos\theta$ ,

mg\sin\theta-\mu mg\cos\theta=ma

a=g(\sin\theta-\mu\cos\theta)

If the sled travels at acceleration $a$ for time $t$ starting from rest, the distance it covers is

d=\frac{1}{2}at^2=\frac{gt^2}{2}(\sin\theta-\mu\cos\theta)

Solving for $\mu$ ,

\frac{2d}{gt^2}=\sin\theta-\mu\cos\theta

\mu=\frac{1}{\cos\theta}\left(\sin\theta-\frac{2d}{gt^2}\right)=\frac{1}{\cos 25^{\circ}}\left(\sin 25^{\circ}-\frac{2(85\,\mathrm{m})}{(10\,\mathrm{m/s^2})(17\,\mathrm{s})^2}\right)=0.40

so the answer is B.