A 25-kg block is connected to a 30-kg block by a light string that passes over a

Question

A 25-kg block is connected to a 30-kg block by a light string that passes over a frictionless pulley. The 30-kg block is connected to a light spring of force constant 200 N/m, as in the figure below. The spring is unstretched when the system is as shown in the figure, and the incline is smooth. The 25-kg block is pulled 15 cm down the incline (so that the 30-kg block is 35 cm above the floor) and is released from rest. Find the speed of each block when the 30-kg block is 20 cm above the floor (that is, when the spring is unstretched). m/s

A 25-kg block is connected to a 30-kg block by a light string that passes over a frictionless pulley. The 30-kg block is connected to a light spring of force constant 200 N/m, as in the figure below. The spring is unstretched when the system is as shown in the figure, and the incline is smooth. The 25-kg block is pulled 15 cm down the incline (so that the 30-kg block is 35 cm above the floor) and is released from rest. Find the speed of each block when the 30-kg block is 20 cm above the floor (that is, when the spring is unstretched). m/s

Explanation / Answer

At the point when the block is moved down, just before release, the following forces exist:

F of the 30kg block = mg = 30*9.8 N downwards
F of the spring = 195*0.15 N downwards

F of the 25kg block (a bit more complicated) = sin(40)*mg = sin(40)*25*9.8 (only the vertically downward component has effect)

So the net force in the system is
F = 30*9.8 + 195*0.15 - sin(40)*25*9.8 = 165.8N (downwards)

The acceleration, a is
a = F/m where m = the combined masses
a = 165.8/(25+30) = 3.014 m/s/s

v^2 = u^2 + 2as
v = sqrt(0 + 2*3.014*0.15)
v = 0.95 m/s

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