The triangular block is moving horizontally with constant velocity $V_{0}$ as shown in the figure. The block A always remain stationary with respect to triangular block. In time t,

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The work done normal reaction on block A is $W_{2} = \frac{m g}{2} v_{0} t \sin (2 θ)$

The work done by normal reaction on triangular block is zero

The work done by normal reaction (between block A and the triangular block) is zero
on the system (triangular block + block A) in any frame of reference

The work done by normal reaction on triangular block is $W_{1} = \frac{- m g}{2} v_{0} t \sin (2 θ)$

answer is A, C, D.

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Detailed Solution

(ABCD)
SOL.: As the block a always remains stationary with respect to triangular block
$f = m g \sin θ N = m g \cos θ$
Question Image
Displacement of the triangular block w.r.t ground = $v_{0} t$
Work done by normal reaction $W_{1} = - N v_{0} t \cos (90^{0} - θ)$
$W_{1} = m g \cos θ t \cos (π / 2 - θ)$
$= - m g \cos θ \sin θ v_{0} t = \frac{- m g v_{0} t \sin 2 θ}{2}$
B) If normal reaction is internal force of a system, work done normal reaction on the system is zero in all frames of reference
C) $\begin{array}{l} W_{1} + W_{2} = 0 \\ W_{2} = - W_{1} = \frac{m g v_{0} t \sin 2 θ}{2} \end{array}$
D) $W_{1} + W_{2} = 0$