Pulling on a cylinder and energy lost to friction

[ThisIsWhyImABioMajor]

A spool of thin wire (with inner radius r = 0.50 m, outer radius R = 0.65 m, and moment of inertia Icm = 1.06 kg*m2 pivots on a shaft. The wire is pulled down by a mass M = 1.35 kg. After falling a distance D = 0.54 m, starting from rest, the mass has a speed of v = 70.200 cm/s. Calculate the energy lost to friction during that time.
Not sure where even to begin!

 

[vanesch]

Welcome to PF !

Please read our guidelines concerning homework which you agreed upon:
- post them in the appropriate homework section (in this case: intro physics; I'll move the thread after this post)

- show some work before others are even allowed to help you.

This policy is not to make life hard for you, on the contrary...

However, I'm going to give you a hint in approaching the problem:
if you know the energy at the beginning, and at the end, then the difference might have something to do with friction (because if there weren't any, both would be the same).

 

[kyle8921]

I would approach this problem by first identifying the relevant equations and variables. In this case, the key equation is the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. The variables in this equation are the work (W), the force (F) applied to the object, the displacement (d) of the object, and its final and initial kinetic energies (Kf and Ki).

In this scenario, the force applied to the spool of wire is the weight of the mass (Mg), and the displacement is the distance it falls (D). Therefore, the work done on the spool is given by W = Fd = (Mg)(D). The initial kinetic energy of the mass is zero, since it starts from rest, and its final kinetic energy is given by Kf = 1/2*M*v^2, where v is the final speed of the mass.

Using the work-energy theorem, we can now solve for the energy lost to friction during this process. The change in kinetic energy is Kf - Ki = 1/2*M*v^2 - 0 = 1/2*M*v^2. Therefore, the energy lost to friction is given by:

Efriction = W - (Kf - Ki) = (Mg)(D) - 1/2*M*v^2

Substituting in the given values, we get:

Efriction = (1.35 kg)(9.8 m/s^2)(0.54 m) - 1/2*(1.35 kg)*(0.702 m/s)^2 = 6.599 J

Therefore, the energy lost to friction during the process of pulling the wire down by a mass of 1.35 kg and causing it to reach a speed of 70.200 cm/s is approximately 6.599 J. This represents the amount of energy that was converted into heat due to friction between the wire and the shaft during this process.