A light spring of force constant 3.65 N/m is compressed by 8.00 cm and held betw

Question

A light spring of force constant 3.65 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.590 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 0.590 kg block
0.000


Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

m/s
0.100


Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s

m/s
0.475

m/s

m/s

Explanation / Answer

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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