Amass m1 = 3.5 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.8 kg which hangs freely from the string. when released, the hanging mass falls a distance d = 0.71 m. 1) how much work is done by gravity on the two block system? 34 j (all of my answers are correct up to the last question, which i need on) 2) 2) how much work is done by the normal force on m1? 0 j 3) what is the final speed of the two blocks? 2.84 m/s 4) how much work is done by tension on m1? 14.1 j 5) what is the tension in the string as the block falls? 19.9 n 7) what is the net work done on m2? this is the question i need on. i gave the answers to the previous questions because something in there might be needed for solving this part of the problem.

1) 33.4 J 
In the described system, the only available energy is the potential gravitational energy for mass m2. So: 0.71 m * 9.8 m/s^2 * 4.8 kg = 33.3984 kg*m^2/s^2 = 33.3984 J 
 2) 0 J
 Since the table is frictionless and the normal force is towards the table's surface, there is no movement or work done via the normal force on m1. So 0 joules of work is done. 
 3) 2.84 m/s
 Using the equation E = 0.5MV^2 and the knowledge that E = 33.3984 J from 1 above, we get
 33.3984 kg*m^2/s^2 = 0.5*(3.5 kg + 4.8 kg)V^2
 33.3984 kg*m^2/s^2 = 0.5*8.3 kg V^2
 33.3984 kg*m^2/s^2 = 4.15 kg V^2
 8.047807229 m^2/s^2 = V^2
 2.83686574 m/s = V 
 4) 14.1 J
 Using the equation E = 0.5MV^2 and the known velocity, we get
 E = 0.5MV^2
 E = 0.5*3.5kg* (2.84 m/s)^2
 E = 0.5*3.5kg* 8.0656 m^2/s^2
 E = 14.1148 kg*m^2/s^2
 E = 14.1 J 
 5) 19.9 N
 The tension in the string is the force needed to perform 14.1 J over a distance of 0.71 m. So:
 14.1 J / 0.71 m = 19.88 J/m = 19.88 kg*m/s^2 = 19.88 N 
 7) 19.3 J
 The NET work is the GROSS work performed on m2 minus the work that m2 performed. The gross work is given by 1 above (33.4 J) minus the work performed on m1 given by 4 above (14.1 J). So the net work is 33.4 - 14.1 =
19.3 J
 You can also calculate the net work performed on m2 using the equation E = 0.5MV^2, so
 E = 0.5MV^2
 E = 0.5*4.8 kg * (2.84 m/s)^2
 E = 0.5*4.8 kg * 8.0656 m^2/s^2
 E = 19.35744 kg*m^2/s^2
 E = 19.4 J
 And as you can see, the kinetic energy that m2 has matches the NET work performed on m2, which makes sense since energy can be neither created, nor destroyed.