Question

A rigid body of mass 0.3 kg is taken slowly up an inclined plane of length 10 m and height 5 m (assuming the applied force to be parallel to the inclined plane), and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. using g = 9.8 $m/s^2$ find the
(a) work done y the gravitational force over the round trip
(b) work done by the applied force over the upward journey
(c) work done by frictional force over the round trip
(d) kinetic energy of the body at the end of the trip?

Answer 1

It is given that

$m=0.3Kg$

$h=10m$

$l=5m$

$\mu =0.15$

(a) As the displacement over the round trip is zero so, work done by the gravitational force over the round trip is zero.

(b) work done by the applied force over the upward journey

$W_{up}=F\times d$

$=(mg\sin \theta +f)h$

$=mg(\sin \theta +\mu \cos \theta )h$

$=0.3\times 9.8\times 10\left(\right(\frac{5}{10})+0.15(\frac{\sqrt{75}}{10}\left)\right)$

$=29.4(0.5+0.129)$

$=18.51J$

(c) work done by frictional force over the round trip

$W=2fh$

$=2\left(\mu mgh\cos \theta \right)$

$=2\times 0.15\times 0.3\times 9.8\times \frac{\sqrt{75}}{10}\times 10$

$=7.638J$

(d) kinetic energy of the body at the end of the trip

$K=\frac{1}{2}m{v}^2$

From newton’s second law

$mg\sin \theta -f=ma$

$mg\sin \theta -\mu mg\cos \theta =ma$

$a=g(\sin \theta -\mu \cos \theta )$

$a=9.8(0.5-0.15\times \frac{\sqrt{75}}{10})$

$a=3.62m{s}^{-2}$

$v^2-u^2=2as$

$v^2=72.4$

$v=8.5m/s$

$K=\frac{1}{2}\times 0.3\times 72.4K=10.86J$