How Is Total Work Calculated for Two Connected Blocks with Friction?

[james brug]

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.

Find the total work done on the 20 N block if $$\mu _s\;$$=(coeff. of static friction)=0.500 and $$\mu _k \;\;\;\;$$=(coeff. of kinetic friction)=0.325 between the table and the 20 N block.

Homework Equations

$$w=f\cdot d$$
$$f_k=\mu_k\cdot n$$
$$f_s\leq\mu_s\cdot n$$

The Attempt at a Solution

w=[7.5N-(.325)(20N)](.75m)=.75J--wrong.

 

[james brug]

Never mind. I've solved it myself through some careful research and considerable effort. It is unfortunate that no one was able to answer this in time. Perhaps you people want some monetary compensation? Or maybe no one liked my problem. Not particularly hard, is it?

 

[nlightner]

I would first clarify any potential misunderstandings or ambiguities in the given information. For example, I would confirm whether the 20.0 N and 12.0 N refer to the masses of the blocks or the forces acting on them. I would also clarify the direction of motion for each block and the orientation of the pulley.

Assuming that the given information is correct, I would approach the problem by first calculating the forces acting on each block. For the 20.0 N block, there are three forces to consider: gravity (20.0 N downward), tension in the string (unknown), and friction (unknown). For the 12.0 N block, there are two forces: gravity (12.0 N downward) and tension (unknown).

Next, I would use Newton's second law (F=ma) to determine the acceleration of each block. Since the blocks are connected by a string and moving in opposite directions, their accelerations must be equal in magnitude but opposite in direction.

Once the accelerations are known, I would use the equations for work (W=Fd) and friction (F=μN) to calculate the work done on the 20.0 N block. The work done by friction would depend on whether the block is moving or stationary, so both the kinetic and static coefficients of friction would need to be considered.

In summary, the total work done on the 20.0 N block in this scenario cannot be accurately determined without more information and calculations. I would also consider factors such as the materials and surfaces involved, the angle of the string, and any potential energy changes.