Question

A box of mass 20.0 [kg] slides down from the

top of a surface inclined at 50.0◦ with height of 10.0 [m],

as shown in the figure. The coefficient of kinetic friction

between the box and the inclined surface is µk = 0.350.

When the box reaches the bottom of the incline, what is

the work done by the weight of the box?

Answer 1

$W=A_1+A_2=mgh- F_{fr}l=mgh-\mu N\frac h{\sin\alpha}=mgh-\mu mg\cos\alpha\frac h{\sin\alpha}=mgh(1-\mu\cot \alpha),$

$W=20\cdot 9.8\cdot 10\cdot(1-0.35\cdot\cot 50°)=1384~J.$