Question

A rigid body of mass 3.0 kg is taken slowly up an inclined plane of length 10 m and height 5 m, and then allowed to slide down to the bottom again. The coefficient of friction between body and the plane is 0.15. Using g = 9.8 $m/s^2$. find the
(a) work done by the gravitational force over the round trip
(b) work done by the applied force (assuming it to be parallel to the inclined plane) over the upward journey.

Answer 1

Given

$m=0.3kg$

$h=10m$

$P=5m$

$\mu =0.15$

(a) Work done by gravitational force over the round trip$=0$ [As displacement is zero

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

$W_{up}=Force\times Displacement$

$=(mg\sin \theta +f)h$ [As the block moves up friction would be downward]

$=(mg\sin \theta +\mu mg\cos \theta )h$ [Since $f=\mu N=\mu mg\cos \theta$]

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

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

$=3(5+0.15\sqrt{75})$

$=18.89J$