Question

A 2.5-kg block is sliding along a rough horizontal surface and
collides with a horizontal spring whose spring constant is 320 N/m.
Unstretched, the spring is 20.0 cm long. The block causes the
spring to compress to a length of 12.5 cm as the block temporarily
comes to rest. The coefficient of kinetic friction between the
block and the horizontal surface is 0.25. a) How much work is done
by the spring as it brings the block to rest? b) How much work is
done on the block by friction as the block is in contact with the
spring? c) What was the speed of the block when it first came into
contact with the spring? Show all work. **Diagram
here:** http://imgur.com/T5uu2Ad

Answer #1

The block has a mass of 2.5 kg. At the surface of the earth, the
acceleration of gravity is approximately 9.8 m/s^2, so the block
exerts a downward force of 2.5 x 9.8 = 24.5 newtons. The
coefficient of friction is 0.25, so the horizontal force opposing
the motion of the block = 24.5 x .25 = 6.125 newtons. Since work =
force x distance, the friction work = 6.125 * .075 = 0.46
Joules.

Total work done = 0.9 + 0.46 = 1.36 joules. This equals the energy
of the block at the start of the contact with the spring. The
energy of a moving block = (1/2) m v^2 where m is the mass of the
block and v is the velocity. In this case, the mass = 2.5 kg, so
the equation to be solved is

1.36 = 1.25 v^2

v^2 = 1.088

v = 1.04 m/s

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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...

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