A body slides from rest down an incline of length l = 30 m…

A body slides from rest down an incline of length $l = 30\ \mathrm{m}$ and angle $\theta = 30^\circ$. The coefficient of friction is $\mu = 0.17$. (a) How long does it take to slide the whole length? (b) What is its speed at the bottom?

Explanation of Method

Sliding down a 30 m incline with friction.

Step 1: Acceleration

\[ a = g\,(\sin\theta - \mu\cos\theta), \]

\[ a = 9.81\,(0.5 - 0.17 \cdot 0.866). \]

\[ a \approx 3.46\ \mathrm{m/s^2}. \]

Step 2: (a) Time

\[ t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{60}{3.461}}. \]

\[t \approx 4.16\ \mathrm{s}\]

Step 3: (b) Final speed

\[ v = \sqrt{2al} = \sqrt{2 \cdot 3.461 \cdot 30}. \]

\[v \approx 14.4\ \mathrm{m/s}\]

Check

$ v = a t = 3.461 \cdot 4.164 \approx 14.4\ \mathrm{m/s}\ \checkmark $; average speed $ 30/4.16 = 7.2 = v/2\ \checkmark $

$30^\circ$ $mg$ $N$ $\mu N$ $a = 3.46\ \mathrm{m/s^2}$ $l = 30\ \mathrm{m}$

The slope component $mg\sin\theta$ pulls the body down, friction $\mu mg\cos\theta$ holds it back; their difference gives $a = g(\sin\theta - \mu\cos\theta)$.

Result

\[\boxed{\,t \approx 4.16\ \mathrm{s}\,}\]

\[\boxed{\,v \approx 14.4\ \mathrm{m/s}\,}\]

Remark

About the original solution. The formula is right, but it evaluated $a \approx 3.56\ \mathrm{m/s^2}$ instead of 3.46, and then gave $t \approx 4.61\ \mathrm{s}$, which does not even follow from its own 3.56 (that would be 4.11 s). The correct values are $t \approx 4.16\ \mathrm{s}$ and $v \approx 14.4\ \mathrm{m/s}$ (not 14.63).