Question

an inclined plane making an angle of 25 degree with the horizontal has a pulley at its top. A 30kg block on plane is connected to a freely hanging 20kg block by means of a string passing over the pulley. Compute the distance the 20kg block will fall in 2 second starting from rest. Neglect the friction.

Answer 1

Let

M$_A$ = 30 kg

M$_B$ = 20 kg

For Block B

$M_Bg-T=M_ba$

$20×9.8-T=20a$

$T+20a=196.......i)$

For Block A

$T-M_Agsin25=M_aa$

$T-30×9.8×sin25=30a$a

$T-30a=124.25......ii)$

$i-ii$

$50a=196-124.25$

$a=1.435m/s^2$

So distance traveled by 20 kg block is

$Y=V_{oy}t+\frac{1}{2}at^2=0+\frac{1}{2}×1.435×2^2$

$Y=2.87m$