A rigid body of mass 0.3 kg is taken . . .

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/s², find the 

(a) work done by the gravitational force over the round trip. 
(b) work done by the applied force over the upward journey. 
(c) work done by the frictional force over the round trip. 
(d) kinetic energy of the body at the end of the trip. 

Doubt by Krish 

Solution : 

mass (m) = 0.3 kg
Length of inclined plane (H) = 10 m
Height of the inclined plane (P) = 5 m 
Coefficient of friction (µ) = 0.15
Angle of inclination of the inclined plane = θ
sinθ=P/H
sinθ=5/10
sinθ=1/2
sinθ=sin30°
θ=30°

a) Work done by the gravitational force over the round trip
W=mg(h2-h1)
W=mh(0)
W=0

b) Work done by the applied force over the upward journey. 
W=mg(sinθ+µcosθ)×H
substituting the value you will get 
W=18.5 J

c) Work done by the frictional force over the round trip. 

Work done by the frictional force in going up = work done by the frictional force in going down

Work done by the frictional force over the round trip = 2×work done by the frictional force in going up 
= 2×fScos[Angle]
= 2×fScos180°
= -2fS
=-2(fH)
=-2µRH
=-2µmgcosθH
=-2(0.15)(0.3)(9.8)(√3/2)(10)
= -(0.15)(0.3)(9.8)(√3)(10)
=-7.6 J

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

As per the work energy theorem 
Net Work Done = Change in KE
Work done by Gravity + Work done in moving the body up + Work done by friction = Final KE-Initial KE
0+18.5+(-7.6)=Kf-(0)
10.9=Kf
Kf=10.9 J