A mass m1 = 4.9 kg rests on a frictionless table and connected by a massless str

Question

A mass m1 = 4.9 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.6 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.79 m.

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

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

3)What is the final speed of the two blocks?

4)How much work is done by tension on m1?

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

6) What is the NET work done on m2?

Explanation / Answer

Part A)

W = Fd

W = (4.6)(9.8)(.79)

W = 35.6 J

Part B)

The normal force never does any work

W = 0 J

Part C)

For the hanging block, the sum of forces is mg - T = ma

T = (4.6)(9.8) - 4.6a

For the block on the table...

T = ma = 4.9a

Set the T's equal

4.9a = (4.6)(9.8) - 4.6a

a = 4.74 m/s2

Then apply vf2 = vo2 + 2ad

vf2 = 0 + 2(4.74)(.79)

vf = 2.74 m/s

Part D)

W = Td

W = (4.9)(4.74)(.79)

W = 18.4 J

Part E)

T = (4.9)(4.74) = 23.3 N

Part F)

Net work = (mg-T)(d)

W = [(4.6)(9.8) - 23.3](.79)

W = 17.2 J

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