The coeffcient of friction between the 7.9 kg mass and the table is 0.55, and the coeffcient of friction between the 5 kg mass and the table is 0.38.

Consider the motion of the 37 kg mass
which descends by the amount of y = 0.45m after releasing the system from rest. The acceleration of gravity is 9.8 m/s2 : Find the work done against friction. Answer in units of J.
part2 Find the speed of the 37 kg mass. Answer in units of m/s.
part3
Find the time it takes for the 37 kg mass to fall a distance of y = 0.45 m.

Diagram: mass one on the left edge of table, mass 2 on middle of table and mass three handing down on a pulley on the right hand side of the table all connected by a string.

I am currently taking physics myself, so I'm not certain, but I can at least get you started on part 1.
Work = (Force) * (distance)
In this case, the force is friction, and the distance is .45m (if the 37kg mass falls .45m, the weights on table are pulled the same distance horizontally).
Friction force is (coeff. of friction)*(normal force), and the normal force is the same as the weight for a horizontal surface, which is (mass)*(grav. accel). If you work in reverse order of these steps, you should be able to find the work in J (which are the same as kg*m^2/s^2). Don't forget to add the friction forces for both masses. Hope this helps.

Nice question

To solve part 1, we need to find the work done against friction.

Work is defined as the force applied in the direction of the displacement. In this case, the force is the friction force and the displacement is the distance the 37 kg mass descends, which is given as y = 0.45 m.

To calculate the friction force, we multiply the coefficient of friction by the normal force. The normal force for each mass is equal to the weight (mass * gravitational acceleration).

Let's calculate the normal forces for each mass:
Normal force for the 7.9 kg mass = (7.9 kg) * (9.8 m/s^2) = 77.42 N
Normal force for the 5 kg mass = (5 kg) * (9.8 m/s^2) = 49 N

Now, let's calculate the friction forces for each mass:
Friction force for the 7.9 kg mass = (0.55) * (77.42 N) = 42.58 N
Friction force for the 5 kg mass = (0.38) * (49 N) = 18.62 N

Since both masses are connected by a string, the work done against friction for each mass is the friction force multiplied by the distance y:
Work done against friction for the 7.9 kg mass = (42.58 N) * (0.45 m) = 19.16 J
Work done against friction for the 5 kg mass = (18.62 N) * (0.45 m) = 8.38 J

To find the total work done against friction, we add the work done against friction for both masses:
Total work done against friction = 19.16 J + 8.38 J = 27.54 J

Therefore, the answer to part 1 is 27.54 J.