Mass-spring system with friction

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A mass M slides on a table, compressing a spring upon collision, which allows for the determination of the spring constant k. The discussion explores using the work-energy principle to derive the coefficient of kinetic friction μk in terms of the compression distance l, initial speed v0, and gravitational acceleration g. It also analyzes the energy relations during half the cycle, leading to a connection between k, l, v0, μk, and g. The work done by friction is debated, with suggestions that it could be calculated based on the distance traveled during compression and rebound. The conversation emphasizes the importance of integrating these formulas to solve for k effectively.
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A mass M slides across a horizontal table. It collides with a spring, compresses the spring, and then the mass-spring system rebounds. This system can be used to find the spring constant k. When the mass first hits the spring at x = 0, it has speed v0.

a.) Let the coefficient of kinetic friction be μk. Assume that the spring plus mass compresses to a distance l, rebounds, and stops when it returns to x = 0, having compressed the spring only once. Use the work-energy relation, Wnon-cons = ΔPE + ΔKE to find the required coefficient of friction μk in terms of l, v0 and g.

b.) Next, consider just half the cycle, with the mass starting out with speed v0 at position x = 0, and stopping (for an instant) at x = l. Use the work-energy relation once again to find a relation between k, l, v0, μk, and g.

c.) Use the results of parts (a) and (b) to solve for the spring constant k in terms of v0, μk, and g.
 
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How would you think to go about solving it other than putting it here for someone else to solve?
 
Yes, very interesting.
Any ideas? Formulas you might be able to use?
 
Well i know that KE is equal to 1/mv^2 and the potential energy of a spring is equal to 1/2kl^2. The work done by friction is going to be equal to μkmgl
 
Won't the work done by friction be 2*μk*m*g*l ? (Twice the distance - to compression and back to equilibrium.)

So maybe add it all up and solve for μk?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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