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Question

A light spring initially compressed 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
j

Answer 2

Ok, here is how i think you do it, first you get the energies equation:

1/2ka2 + 1/2 kb2 + Cmgb = 1/2mVf2 + 1/2mVo2

where a = 20 cm, b = 5 cm, C= 0.6, k the constant of the spring, Vf = 0 and Vo = 1.5 m/s

now, here is how you solve the equation:

1/2k(a2 + b2) + Cmgb = 1/2 mVo2

k = 2*m*(Vo2-2Cgb)/(a2+b2)

K= 2*0.3kg*((1.5m/s)2-2*0.6*9.81m/s2*0.05m)/((0.02m)2+(0.05m)2)

k = 23.5 N/m