A 20.0 kg block is connected to a 30.0 kg block by a string that passes over a l

Question

A 20.0 kg block is connected to a 30.0 kg block by a string that passes over a light, frictionless pulley. The 30.0 kg block is connected to a spring that has negligible mass and a force constant of 220 N/m, as shown in the figure below. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0 kg block is pulled 24.0 cm down the incline (so that the 30.0 kg block is 44.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0 kg block is 20.0 cm above the floor (that is, when the spring is unstretched).
m/s

Explanation / Answer

The energy in the spring
.5*k*d^2
plus the loss of PE of the hanging block
30*g*d
less the gain in PE of the block on the incline
20*g*sin40*d
equals the KE of the 30 k and 20 k blocks
.5*50*v^2

solve for v (they are the same because of the massless, frictionless and no-stretch string that connects them)

.5*k*d^2+30*g*d-20*g*sin40*d=.5*50*v^2


v=sqrt(2*(.5*220*0.24^2+30*9.81*0.24-
20*9.81*sin40*0.24)/50)

v=1.367 m/s

Both the blocks are moving with the same velocity i.e 1.367 m/s

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