Question

Amanpushingacrateofmassm592.0kgataspeedof v 5 0.850 m/s encounters a rough horizontal surface of length , 5 0.65 m as in Figure P5.18. If the coefficient of kinetic fric- tion between the crate and rough surface is 0.358 and he exerts a constant horizontal force of 275 N on the crate, find (a) the magnitude and direction of the net force on the crate while it is on the rough surface, (b) the net work done on the crate while it is on the rough surface, and (c) the speed of the crate when it reaches the end of the rough surface.

Answer 1

(a) The net force

$F_{net}=F-F_f=F-\mu mg=275-0.358\cdot 92\cdot 9.8=-47.8N$

Opposite to the motion

(b) The net work done

$W_{net}=F_{net}\cdot \Delta x=-47.8\cdot 0.65=-31.1 J$

(c) From the Work-Energy Theorem we have

$W=\frac{1}{2}(mv^2_f-mv^2_i)$

$v_f=\sqrt{ \frac{2W+mv^2_i}{m}}=\sqrt{ \frac{2\cdot (-31.1)+92 \cdot 0.85^2}{92}}\approx0.22 m/s$