Help: Kinetic Energy and Friction - A block pulled by another on a pulley

[aarno]

Help: Kinetic Energy and Friction -- A block pulled by another on a pulley

In the system the following diagram shows, the block M (mass of 15.65 kg) is initially moving to the left with a speed of 2.26 m/s. The mass of m is 8.26 kg and the coefficients of friction are μs = 0.411 and μk = 0.304. The string is massless and the pulley is massless and frictionless. How fast (m/s) will M be traveling when m has fallen through a height h=2.47 meters?

diagram: http://ce.byu.edu/courses/univ/694820121008/media/8_problem_blockmass.jpg

 

[Doc Al]

Well, what do you think? What have you done so far?

 

[aarno]

^

The pulley makes it so that the second mass exerts a force purely in the x direction, not affecting the normal force. I used left (or down, for the second mass), as the positive direction.
The force exerted by mass m is equal to its mass times gravity, and delta-x equals 2.47 as provided in the problem. Therefore,
W = F * delta-x = (8.26)(9.8)(2.47) = 199.942

The normal force is equal to the weight of mass M, so the force of kinetic friction:
fk = m*g*μk = (15.65)(9.8)(0.304) = 46.6245

Using the following equation, then plugging in the values:
0.5*m*vf^2 = 0.5*m*vf^2 - fk*d + W
0.5*15.65*vf^2 = 0.5*15.65*2.26^2 - 46.6245*2.47 + 199.942

This give vf = 3.993 m/s, while the correct response is 3.49 m/s.

 

[Doc Al]

The force exerted by mass m is equal to its mass times gravity,

That's incorrect. Note that the masses are accelerating. You'll have to solve for the tension in the rope.

 

[rwisz]

To solve this problem, we can use the principle of conservation of energy. Initially, the total energy of the system is in the form of kinetic energy, as the block M is moving with a speed of 2.26 m/s. As the block m falls through a height of 2.47 meters, some of this energy will be converted into potential energy. However, due to the presence of friction, some of the energy will also be lost in the form of heat.

To calculate the final speed of block M, we can use the equation for conservation of energy:

Initial kinetic energy = Final kinetic energy + Final potential energy + Work done by friction

The initial kinetic energy of block M can be calculated as 1/2 * mass * velocity^2 = 1/2 * 15.65 kg * (2.26 m/s)^2 = 40.39 J

The final potential energy of block m can be calculated as mass * gravity * height = 8.26 kg * 9.8 m/s^2 * 2.47 m = 202.97 J

The work done by friction can be calculated as μk * mass * gravity * distance = 0.304 * 8.26 kg * 9.8 m/s^2 * 2.47 m = 60.04 J

Substituting these values into the conservation of energy equation, we get:

40.39 J = 1/2 * 15.65 kg * v^2 + 202.97 J + 60.04 J

Solving for v, we get v = 3.04 m/s.

Therefore, the final speed of block M will be 3.04 m/s when block m has fallen through a height of 2.47 meters.