In the arrangement shown, by what acceleration the boy must move up so that 100 kg block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere. ( $g = 10 ms^{- 2}$ )

Question

In the arrangement shown, by what acceleration the boy must move up so that 100 kg block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere. ( g = 10 ms − 2 )

TestHub · Accepted Answer

Block (100 kg) stationary on the smooth incline: the rope tension balances the weight component along the incline, $T = 100 g sin 5 3^{\circ} = 100 (10) (0.8) = 800 N$ .

Boy (50 kg) hangs on the same rope and moves up with acceleration $a$ : $T - 50 g = 50 a \Rightarrow 800 - 500 = 50 a \Rightarrow a = 6 m/ s^{2}$ .

Physics - NLM Question with Solution | TestHub

Options:

Doubts & Discussion