Chapter 7 Problem 64

Problem

Assume the coefficient of friction is $\mu _{k}=0.10$ on the incline.

(a) How much work is done by the horizontal force $F_{P}=150$ N on the 18-kg block when the force pushes the block 5.0 m up along the 32° incline?

(b) How much work is done by the gravitational force on the block during this displacement?

(c) How much work is done by the normal force?

(d) What is the speed of the block (assume that it is zero initially) after this displacement?

[Hint: Work-energy involves net work done.]

Solution

$\sum F_{y}=F_{N}-F_{p}\sin \phi -mg\cos \phi =0$ $\rightarrow F_{N}=F_{P}\sin \phi +mg\cos \phi$

$F_{f}=\mu _{k}F_{N}=\mu _{k}\left(F_{P}\sin \phi +mg\cos \phi \right)$

(a)

The angle between the pushing force and the displacement is 32°

$W_{P}=F_{P}d\cos \theta =\left(150{\text{N}}\right)\left(5.0{\text{m}}\right)\cos 32^{\circ }=636.0{\text{J}}\approx 640{\text{J}}$

(b)

The angle between the pushing force and the displacement is 122° $W_{P}=F_{G}d\cos \theta =mgd\cos \theta =\left(18{\text{kg}}\right)\left(9.80{\text{m/s}}^{2}\right)\left(5.0{\text{m}}\right)\cos 122^{\circ }=-467.4{\text{J}}\approx -470{\text{J}}$

(c)

Because the normal force is perpendicular to the displacement, the work done by the normal force is 0

(d)

To find the net work, we need the work done by the friction force. The angle between the friction force and the displacement is 180° $W_{f}=F_{f}d\cos \theta =\mu _{k}\left(F_{P}\sin \phi +mg\cos \phi \right)d\cos \theta$ $=(0.10)\left(\left(150{\text{N}}\right)\sin 32^{\circ }+\left(18{\text{kg}}\right)\left(9.80{\text{m/s}}^{2}\right)\cos 32^{\circ }\right)\left(5.0{\text{m}}\right)\cos 180^{\circ }$ $=-114.5{\text{J}}$

$W=W_{P}+W_{g}+W_{N}+W_{f}=\Delta K={\frac {1}{2}}mv_{f}^{2}-{\frac {1}{2}}mv_{i}^{2}\rightarrow$

$v_{f}={\sqrt {\frac {2W}{m}}}$ $={\sqrt {\frac {2(636{\text{J}}-467.4{\text{J}}-114.5{\text{J}}}{\left(18{\text{kg}}\right)}}}=2.5{\text{m/s}}$