Please solve this physics porblems

Question

A mass m₁ = 5.6 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m₂ = 3 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.73 m.

1)

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

J 

2)

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

J 

3)

What is the final speed of the two blocks?

m/s 

4)

How much work is done by tension on m₁?

J 

5)

What is the tension in the string as the block falls?

N 

6)

What is the NET work done on m₂?

Answer 1

1. The work done by gravity is given by the equation w=m₂*g*d or the mass times acceleration due to gravity (force) times the distance the block travels (force times distance is equal to work) so w=3kg*9.81m/s²*0.73m or 21.5J
2. If you draw a free body diagram and map the forces acting on the blocks you'll see that the normal force is completely vertical, since the block doesn't travel in a vertical direction the normal force isn't doing any work. Remember W = F x D
3. Here we can use the relationship between work and energy. In this case m₂​ *g *d = 1/2*​(m1​+m2​)*v² Now we can just plug in our values and solve the equation for v. That gives us 2.23m/s
4. Here we can also use the relationship between work and energy, so the work done by the tension on the rope is equal to the change in energy when the block drops. So W = 1/2*m₁*v² since we already have mass one and the velocity (the one we got in the last problem) we can just plug and chug. 14J
5. If you draw free body diagrams and apply Newton’s 2nd law you’ll see that for block m1 the only horizontal force is the tension, so T = m1 * a. For block m2 the downward force of gravity minus the tension equals m2 * a, or m2 * g – T = m2 * a. Putting these two equations together gives T = (m1 * m2 * g) / (m1 + m2). Plugging in the numbers: T = (5.6 kg * 3.0 kg * 9.81 m/s^2) / (5.6 + 3.0) = 19.2 N.
6. The net force on m2 is its weight minus the tension, or Fnet = m2 * g – T. Work is just force times distance, so Wnet = (m2 * g – T) * d. Substituting the values we found: (3.0 * 9.81 – 19.2) N * 0.73 m = 7.49 J. Another way to see it is with the work–energy principle: the net work on m2 is equal to its change in kinetic energy, which is 1/2 * m2 * v^2 = 7.49 J, using the velocity from part 3.