Help On Physics Problem?

Question

A light spring initially stretched 20cm has a 300g mass on the end of it. The system is on a rough table top. The coefficient of kinetic friction is 0.60. The masss is initially shoved inward with a speed of 1.5m/s and compresses the spring 5cm before stopping. What is the spring constant?

Answer 1

Use conservation of energy first

the energy of the 300g (.3kg) mass = 1/2 m v^2
=1/2 * .3 * 1.5 = .225

The energy gets transfered to the spring and to the loss due to friction.

The energy in the spring = 1/2 k x^2
where energy = 1/2 k .05^2 (.05 is 5 cm expressed as meters)
so .225 = 1/2 * k .* 05^2 + .3*9.8*.6*.05
(.225/.05 - .3*9.8*.6)*2/.05=k
k=109.44

check 1/2 *109.44 *.05^2 = energy in the spring
=.1368

.3*9.8*.6*.05 is the loss due to friction
=.0882

The spring constant is k=109.44
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