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A block of mass ‘m’ is stationary with respect to a rough wedge as shown in figure. Starting form rest in time ‘t’, $[m = 1 k g, θ = 30^{0}, a = 2 m s^{- 2}, t = 4 \sec]$ .
For magnitude of work done on the block, in Column-I match the answers in Column-II
Column I Column II
A) By gravity        P) 144 J
B) By normal reaction Q) 32 J
C) By friction         R) 48 J
D) By all the forces S) 160 J

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$A \to Q; B \to R; C \to S; D \to P$

$A \to S; B \to P; C \to R; D \to Q$

$A \to Q; B \to R; C \to P; D \to S$

$A \to S; B \to Q; C \to P; D \to R$

answer is B.

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Detailed Solution

$V = a t = 2 \times 4 = 8; S = \frac{1}{2} a t^{2} = 16 m$

$K E = \frac{1}{2} m v^{2} = 32 J$

By gravity $= - m g h = - 1 \times 10 \times 16 = - 160 J$

By normal reaction $N \cos θ S = 144 J$

By friction $f \sin θ S = 48 J$

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