How Is Total Work Calculated for Blocks on a Pulley System with Friction?

[lil_50]

Homework Statement

Two blocks are connected by a very light string passing over a massless and frictionless pulley. The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.

1. Find the total work done on 20.0-N block if u_s=0.500 and u_k=0.325 between the table and the 20.0-N block.
2. Find the total work done on 12.0-N block if u_s=0.500 and u_k=0.325 between the table and the 20.0-N block.

Homework Equations

The Attempt at a Solution

Well the first thing we were supposed to do was find the work done if there was no friction and for the first I got the answer to correctly be 5.625J and the second to be 3.375J. So since the system was already moving I tried to just take the normal force multiplied by the coefficient of kinetic energy but that answer was incorrect. PLEASE HELP.

 

[Doc Al]

So since the system was already moving I tried to just take the normal force multiplied by the coefficient of kinetic energy but that answer was incorrect.

Not sure what you did. Multiplying the normal force times the coefficient of kinetic friction will give you the friction force. What then?

How did you find the answer to part 1?

 

[nlightner]

I would approach this problem by first breaking it down into its components and identifying the forces at play. In this case, we have two blocks connected by a string, and the forces acting on them are gravity, tension from the string, and friction.

To find the total work done on the 20.0-N block, we need to consider the work done by each of these forces. The work done by gravity is easy to calculate, as it is simply the product of the force (20.0 N) and the distance moved (75.0 cm downward). This gives us a value of 15 J.

Next, we need to consider the work done by the tension force. Since the string is light and the pulley is frictionless, the tension force does not do any work on the block. Therefore, the work done by the tension force is 0 J.

Finally, we need to consider the work done by friction. We are given the coefficients of static and kinetic friction, so we can calculate the magnitude of the friction force using the equation F_friction = u_s * N, where N is the normal force. In this case, the normal force is equal to the weight of the block, which is 20.0 N. This means the friction force is 10 N (20.0 N * 0.500). Since the block moves a distance of 75.0 cm to the right, the work done by friction is equal to the force multiplied by the distance, giving us a value of 7.5 J.

Therefore, the total work done on the 20.0-N block is the sum of these three values: 15 J + 0 J + 7.5 J = 22.5 J.

Similarly, for the 12.0-N block, we can calculate the work done by gravity (9 J) and the work done by tension (0 J). However, since the block is moving downward, the friction force is acting in the opposite direction of the motion. This means the work done by friction is negative, so we need to subtract it from the total work done. Using the same calculations as before, we find that the work done by friction is -3.9 J. Therefore, the total work done on the 12.0-N block is 9 J + 0 J - 3.9 J = 5.1 J.

In summary, the