A mass m 1 = 4.9 kg rests on a frictionless table and connected by a massless st

Question

A mass m1 = 4.9 kg rests on a frictionless table and connected by a massless string to another mass m2 = 5.3 kg. A force of magnitude F = 26 N pulls m1 to the left a distance d = 0.88 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

7)

What is the NET work done on m1?

Explanation / Answer

1)
Work = F*s = force*distance = 26*0.88 = 22.88 Nm

2) 0 N, because there is no movement in direction of the normal force

3) v^2 = 2as where a = F/(m1+m2) = 26/10.2 = 2.549 m/s^2
v^2 = 2*2.549*0.88
v = 2.11808 m/s

4) the tension is
T = m2*a = 5.3*2.549 N = 13.5097 N
Work = tension*distance = T*s = 13.5097*0.88 Nm = 11.888536 Nm

5) T = m2*a = 5.3*2.549 N = 13.5097 N

6) there are two forces on m1 that do work: F and T:
work by F = F*s = 26*0.88 Nm
work by T = - 13.5097*0.88 = 11.888536 Nm ( opposite direction of T vs. F)

7) Net work = (F - T)*s = (27 - 13.5097)*0.88 = 11.871464 Nm

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