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.
1 answer
1 answer