A mass m 1 = 5 kg rests on a frictionless table and connected by a massless stri

Question

A mass m1 = 5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 3.6 kg. A force of magnitude F = 40 N pulls m1 to the left a distance d = 0.77 m.

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

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

3)What is the final speed of the two blocks? m/s

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

5)What is the tension in the string? N

6)The net work done by all the forces acting on m1 is:

positive

zero

negative

7)What is the NET work done on m1?

Explanation / Answer

Acceleration of the System => a = F/(m1 + m2) = 4.65 m/s2

1.) Work done by F on system = F.s = 40x0.77 = 30.8 J

2.) Work done by Normal = 0 , because no displacement in the direction of normal force.

3.) Final speed of the blocks = v = sqrt(2as) = 2.68 m/s

4.) Now , Tension , T = m2a = 16.74 N

Work done by Tension = T.s = 16.74X0.77 = 12.88 J

5.) Tension in the String = 16.74 N

6.) Net work done by all force on m1 is positive .

because Fnet = F - T = 40 - 16.74 = 23.26 N towards left , same as the direction of displacement

7.) Net work done on m1 = Fnet xa = 23.26x4.65 = 108.16 J

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