Answer to Question #98307 in Physics for Ravindra

Question

Two blocks are connected by a massless string that runs over the edge of a pulley as shown in (Figure 1). The mass of the pulley is 5 kg . Assume there is no friction between the blocks and the inclined planes.

If the blocks are released from rest, how far would they slide in 2.5 s  seconds?

Answer 1

$$Ma=Mg\sin{25}-T, ma=T-mg\sin{55}$$

$$(M+m)a=Mg\sin{25}-mg\sin{55}=g(M\sin{25}-m\sin{55})$$

$$(8+5)a=9.8(8\sin{25}-5\sin{55})$$

$$a=-0.539\frac{m}{s^2}$$

Negative sign means that the motion is in leftwards direction.

$$d=0.5|a|t^2=0.5(0.539)(2.5)^2=1.7\ m$$