A mass m_1 = 5.3 kg rests on a frictionless table and connected by a massless st

Question

A mass m_1 = 5.3 kg rests on a frictionless table and connected by a massless string to another mass m_2 = 4.4 kg. A force of magnitude F = 36 N pulls m, to the left a distance d = 0.89 m. How much work is done by the force F on the two block system? How much work is done by the normal force on m_1 and m_2? What is the final speed of the two blocks? How much work is done by the tension (in-between the blocks) on block m_2? What is the tension in the string? The net work done by all the forces acting on m_1 is: positive zero negative What is the NET work done on m_1?

Explanation / Answer

1) Work = F*s = force*distance = 36*0.89 = 32.04 Nm

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

3) v^2 = 2as where a = F/(m1+m2) = 36/9.7 = 3.71 m/s^2
v^2 = 2*3.71*0.89
v = 2.56 m/s

4) the tension is
T = m2*a = 4.4*3.71 N = 16.32 N
Work = tension*distance = T*s = 16.32*0.89 Nm = 14.52 Nm

5)  the tension is
T = m2*a = 4.4*3.71 N = 16.32 N
Work = tension*distance = T*s = 16.32*0.89 Nm = 14.52 Nm


6) there are two forces on m1 that do work: F and T:
work by F = F*s = 36*0.89 Nm
work by T = - 16.32*0.86 Nm ( opposite direction of T vs. F)

7) Net work = (F - T)*s = (36 - 16.32)*0.89 = 17.51 Nm

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