A mass m 1 = 4.2 kg rests on a frictionless table and connected by a massless st

Question

A mass m1 = 4.2 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4 kg. A force of magnitude F = 25 N pulls m1 to the left a distance d = 0.72 m.

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

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

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

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

5)What is the tension in the string?

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

A.)positive

B.)zero

C.)negative

7)What is the NET work done on m1?

Explanation / Answer

Ans:-

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

3) v^2 = 2as

where a = F/(m1+m2) = 25/8.2 = 3.05 m/s^2
v^2 = 2*3.05*0.72
v = 4.39 m/s

4) the tension is
T = m2*a = 4*3.05 N = 12.2 N
Work = tension*distance = T*s = 12.2*0.72 Nm = 8.78 Nm

5) see 4)

6) there are two forces on m1 that do work: F and T:
work by F = F*s = 25*0.72 Nm
work by T = - 12.2*0.72 Nm ( opposite direction of T vs. F)
Net work = 9.216Nm ........Positive

7) Net work = (25-12.2)*0.72 = 9.216Nm

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