can you please help me with this? A mass mi = 3.9 kg rests on a frictionless tab

Question

can you please help me with this?

A mass mi = 3.9 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.7 kg. A force of magnitude F = 26 N pulls mi to the left a distance d = 0.85 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 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? The net work done by all the forces acting on m1 is: What is the NET work done on m1?

Explanation / Answer

1. Since surface is frictionless, m1 and m2 move left by F = 26 N force through 0.85 m

Work done = F*d = 26 * 0.85 = 22.1 J

2. Work is F vector dot d vector or Fd*cos(angle)

Here angle between vectors is 0. So, cos 90 = 0

Hence work done is 0 J

3. Final speed of two blocks sqrt (22.1/total mass) = sqrt(22.1*2/8.6) = 2.25 m/s

4. Work done by tension is T*x = m2*a*x

m2= 4.7 kg
x= 0.85 m
a= 26/total mass = 26/8.6 = 3.02
Work done on m2= 3.02*0.85*4.7= 12.06 J

5. Tension is m2*a = 14.19 N

6. Because net force and motion direction are same, net work done on m1 is positive

7. Net work done on m1 is m1*a*x = 3.9*3.02*0.85= 10.01 J

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