A block of mass m = 1 kg is at rest with respect to a rough wedge as shown in figure.
The wedge starts moving up from rest with an acceleration of a = 2m/s 2 and the block remains at rest with respect to wedge then in 4 sec. of motion of wedge work done on block (assume angle of inclination of wedge is θ = 30º and g = 10 m/s 2 ) –
Column I | Column II |
(i) By gravity | [A] 144 J (in magnitude) |
(ii) By normal reaction | [B] 32 J |
(iii) By friction | [C] 160 J |
(iv) By all the forces | [D] 48 J |
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)-[C], (ii)-[A], (iii)-[D], (iv)-[B]
Sol.
N = m (g + a) cos θ
f = m (g + a) sin θ
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