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 co-efficient of friction between the body and the plane is 0.15. Using g = 9.8 m/s 2 find the
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Since the gravitational force is a conservative force therefore the work done in round trip is
zero.
w F = (9.8) (0.3) (1/2) (1 + 0.15 
) (10) J 
– 0.15 × 0.3 × 9.8 × ( 
/2) × 20 J 
– 7.638 J
0.3 × 9.8 × (10/2) (1 – 0.15 × 
) 
10.880 J
Sol. Since, gravitational force is conservative, So, work done by it in round trip is zero.
sin θ = 
= 
⇒ θ = 30º
W F = mg (sin θ + µcos θ ) × λ
= 0.3 × 9.8 
= 18.519 J
W f = –f.s
= –µmg cos θ (2 λ )
= – 2µmg λ cos θ = – 2 × 0.15 × 0.3 × 9.8 × 10 × 
= –7.638 J.
By W.E.T,
K f – K i = W F + W f + W g
K f = (18.519 – 7.638) J = 10.880 J.
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