Question #98307
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?
1
Expert's answer
2019-11-11T15:52:27-0500
Ma=Mgsin25T,ma=Tmgsin55Ma=Mg\sin{25}-T, ma=T-mg\sin{55}

(M+m)a=Mgsin25mgsin55=g(Msin25msin55)(M+m)a=Mg\sin{25}-mg\sin{55}=g(M\sin{25}-m\sin{55})

(8+5)a=9.8(8sin255sin55)(8+5)a=9.8(8\sin{25}-5\sin{55})

a=0.539ms2a=-0.539\frac{m}{s^2}

Negative sign means that the motion is in leftwards direction.


d=0.5at2=0.5(0.539)(2.5)2=1.7 md=0.5|a|t^2=0.5(0.539)(2.5)^2=1.7\ m


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