A mass m 1 = 3.6 kg rests on a frictionless table and connected by a massless st
ID: 1325507 • Letter: A
Question
A mass m1 = 3.6 kg rests on a frictionless table and connected by a massless string to another mass m2 = 3.9 kg. A force of magnitude F = 30 N pulls m1 to the left a distance d = 0.74 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?
J
Explanation / Answer
1) The work done by gravity = m2 * g * h
= 3.9 kg * 9.8 * (0.74 m) = 28.28 J
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2) work done by the normal force on m1 and m2= 0
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3)Use work-energy, W = 1/2 * (m1 + m2) * v2
v = sqrt( 2 * W / (m1 + m2) )
= sqrt (2 * 28.28 / (3.6+3.9)) = 2.75 m/s
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4) The work by tension on m1 = change in kinetic energy of m1
W = 1/2 * m1 * v2
= 1/2 * 3.6 * 2.752= 13.61 J
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5) W = F * d ,
F = W/d = 13.61 / 0.74 = 18.4 N
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6) Work done by tension on m2 is negative since the tension is up and the motion is down
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7) 28.28 J
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