A 2.5-kg block is sliding along a rough horizontal surface and collides with a h

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

Explanation / Answer

Work on a linear spring is (1/2)kd^2 where k is the spring constant and d is the distance the spring moves. In this case, k = 320 N/M, and d = 7.5 cm = .075 meters. So the work on the spring = .5 x 320 x .075 x .075 = 0.9 joules.

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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