Force constant given the coefficient?

Question

A 1.80 kg block slides on a rough horizontal surface. The block hits a spring with a speed of 2.30 m/s and compresses it a distance of 11.0 cm before coming to rest.

If the coefficient of kinetic friction between the block and the surface is 0.560, what is the force constant of the spring?

I'm thinking k=F/X.

K=mass(Potential energy/X)^2

Please help :(

Answer 1

K.E. of the block
=(1/2)mv^2
= (1/2)(1.80)(2.3)^2
= 4.761 J.

This K.E. is utilized in overcoming friction and compressing the spring.
Work done to overcome friction
= μmgx, where x is the distance
= (0.56)(1.8)(9.8)(0.11)
= 1.087 J
Balance of energy goes to compress the spring
= 4.761 - 1.087
= 3.674 J
= (1/2)kx^2
=> k = (2)(3.674) / (0.11)^2 = 607.3 N/m

Answer 2

Just before impact with the spring, TE(x = 0) = KE(x = 0); the total energy of the block when the spring is not compressed (x = 0) is the block's kinetic energy.

Assuming the friction continues while compressing the spring; TE(x = d) = PE(x = d) + WE(d); the total energy when the spring is compressed x = d = .11 m is the potential energy PE(d) = Fd = kdd of the compressed spring; where k is the spring force (F) constant you are looking for, and WE = umgd which is the friction energy lost as the block slides over the rough surface during compression; where u = .560, m = 1.8 kg, and g = 9.81 m/sec^2.

From the conservation of energy TE(x = 0) = KE(0) = 1/2 mv^2 = kdd + umgd = PE(d) + WE(d); where v = 2.3 mps.

Thus 1/2 mv^2 - umgd = kd^2; so that k = m(v^2/2 - ugd)/d^2 and you can do the math. [NB: If no friction is assumed while compressing the spring, just set u = 0 in the above equation and solve for k.]

The physics is the conservation of energy. All the total energy TE is kinetic just as the block strikes the spring. All that kinetic energy is converted into potential energy, which is the spring force F = kd times the distance moved d, and heat energy from friction WE = umgd, which is the friction force times distance moved.