A 17 kg block on a horizontal surface is attached to a horizontal spring of spri

Question

A 17 kg block on a horizontal surface is attached to a horizontal spring of spring constant k = 3.5 kN/m. The block is pulled to the right so that the spring is stretched 8.8 cm beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of 96 N. (a) What is the kinetic energy of the block when it has moved 2.2 cm from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

Explanation / Answer

Part A)

The energy of the strecthed spring is .5kx2

Max energy is (.5)(3500)(.088)2

Energy is 13.6 J

When the spring is released, part of that energy will start to move into KE, as well as some lost to work done by friction (Ffd)

After moving 2.2 cm, the energy total of 13.6 will be .5kx2 + KE + Ffd

13.6 = (.5)(3500)(.066)2 + KE + (96)(.022)

KE = 3.865 J

Part B)

When it slide back to the equilibrium point, we have

Energy max = KE + Ffd

13.6 = KE + (96)(.088)

KE = 5.152 J

Part C)

The maximum KE will be attained when the block first passes through the equilibrium position. That will be when it has its maximum velocity. Every point before or after, will result in a slower velocity. Therefore the Max KE was solved for in part B.

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