Question

A mass m1 = 5.7 kg rests on a frictionless table and connected by a massless string to another mass m2 = 5.8 kg. A force of magnitude F = 44 N pulls m1 to the left a distance d = 0.89 m.

How much work is done by the force F on the two block system?

How much work is done by the normal force on m₁ and m₂?

What is the final speed of the two blocks?

How much work is done by the tension (in-between the blocks) on block m₂?

What is the tension in the string?

What is the NET work done on m₁?

Answer 1

1. Work done by F = 44 x 0.89 = 39.16 Joules

2. Work Done by normal force on m1 and m2 = W = 0 joules since normal force is perpendicular to direction of movement.

3. Final speed v = sqrt( 2 x 39.16/(m1 + m2) )= 6.81^0.5 = 2.61m/s

Hence the final speed of blocks is 2.61 m/s

4. T = F =44N (since horizontal system)

Hence work done by tension = 44 x 0.89 = 39.16 Joules

5. The tension in the string is same as F = 44N

6. The net work done on m1 = 0