Question

if a 100-lb box slides down a skid 20.0 ft long and inclined at an angle of 60° to tue horizontal. at the bottom of the skid the box slides along a level surface of equal roughness. i f the coefficient of kinetic friction of the surface is 0.100, how far will the box travel, and how long a time will elapse after it reaches the level surface before it comes to rest?

Answer 1

From the conservation of energy:

$$mgl\sin{60}=\mu mg L\\l\sin{60}=\mu L\\(20)(0.3048)\sin{60}=0.1 L\\L=52.8\ m$$

Then

$$mgl\sin{60}=0.5mv^2\\gl\sin{60}=0.5v^2\\(9.8)(20)(0.3048)\sin{60}=0.5v^2\\v=10.17\frac{m}{s}$$

$$t=\frac{v}{a}=\frac{v}{\mu g}\\t=\frac{10.17}{0.1\cdot 9.8}=10.4\ s$$