Question

Course Contents » ... » Set 04 (10/09 Tu 10 PM) » Block and a spring A moving 4.40 kg block collides with a horizontal spring whose spring constant is 433 N/m. The block compresses the spring a maximum distance of 13.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.450. What is the work done by the spring in bringing the block to rest? How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? What is the speed of the block when it hits the spring?

Answer 1

Work done by the spring = energy stored in the compressed spring

work = 1/2kx^2

work = 1/2(433 N/m)(.135m)^2

work = 3.9457 J

For friction, Work = Fd, where F is the force of friction which equals 0.45 mg. The distance is .135 m

So W = .45 (4.4 kg)(9.8 m/s^2)(.135 m) = 2.61954 J

This is the answer to the second question.

For the third question:

At the instant of impact the energy is all kinetic (1/2mv^s)

When the spring is compressed, the energy is now split between the work done by friction, .805 J, and the elastic energy of the spring which equals to the work done in compressing the spring which is 1.753 J.

kinetic energy = spring work + friction work

1/2 mv^2 = 3.9457 J +2.61954 J

1/2 mv^2 = 6.56524 J

solving for v

v = sqroot(2 (6.56524 J) / 4.4 kg)

v = 1.7275 m/s