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,

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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) $W_{1} + W_{2} = 0$