A light spring of force constant 3.50 N/m is compressed by 8.00 cm and held betw

Question

A light spring of force constant 3.50 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.450 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is larger than that for kinetic friction. Let the positive direction point to the right.

µk 0.250 kg block 0.450 kg block
0.000 m/s m/s
0.100 m/s m/s
0.494 m/s m/s

Explanation / Answer

Neat explanation for a very similar question.

Question:A light spring of force constant 4.25 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.570 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is larger than that for kinetic friction. Let the positive direction point to the right.

The energy stored in the spring minus energy consumed by friction will equal the total kinetic energy of the two blocks. The momentum before the spring is released is zero, therefore the momentum after the spring is released is zero. This is analogous to a canon firing.
m1V1=m2V2 except V1 and V2 are in opposite directions
m1 = 0.250kg
m2 = 0.570kg = 2.28m1
m1V1=2.28m1V2 => V1 = 2.28V2
The energy stored in the spring is 0.5*4.25*0.08^2 = 13.6 mJ
Without friction
13.6e-3 = (1/2)m1V1^2 + (1/2)m2V2^2 => 0.5*m1(V1^2 + 2.28*V2^2)
13.6e-3 = 0.125[(2.28V2)^2 + 2.28V2^2)] V2 = 0.121m/s V1 = 0.275m/s
These numbers do work in m1V1=m2V2 and (1/2)*4.25*.08^2=0.5[m1V1^2+m2V2^2]
With Friction
Conservation of Momentum must still hold so m1V1=m2V2 => V1=2.28V2 at t = 0
The work done by friction will sum to the energy in the spring = 13.6mJ
The initial velocities will be the same but will quickly decrease to zero due to friction
Later: suddenly occurred to me that there is not enough force in the compressed spring to over come the friction force holding the 0.57kg block in place. That force = 0.57*9.8*0.1 = 0.56N at least.
If F=kx = 4.25*0.08 = 0.34N meaning the larger block would remain stationary and only the 0.25kg block would move. So the energy in the compressed spring would be transfered to the smaller block and 13.6mJ = 0.5*0.25*V^2 => V = 0.33m/s

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