A mass m 1 = 4.6 kg rests on a frictionless table and connected by a massless st

Question

A mass m1 = 4.6 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.1 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.74 m.

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

________________J

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

______________J

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

_______________m/s

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

____________J

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

______________N

6)What is the NET work done on m2?

______________J

Explanation / Answer

Both are connected to a string hence travel same distance.

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

Work done on block 1 is zero

work done on block 2 is mgh = 4.1*9.81*0.74 = 29.76 J

So net work done = 29.76 J

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

Normal force does no work on m1 as the displacement is perpendicular to the force.

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

applying work energy theorem for this system:

29.76 = 0.5*(m1+m2)*v^2

29.76 = 0.5*(4.6+4.1)*v^2

v = 2.62 m/s

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

work done by the tension on m1 = change in kinetic energy of m1 = 0.5 m1 v^2 = 0.5*4.6*2.62^2 = 15.79 J

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

Tension is constant T = m1 m2 g/(m1+m2) = 4.6*4.1*9.81/(4.6+4.1) = 21.266 N

6)What is the NET work done on m2?

Net work done on m2 = change in kinetic energy of m2 = 0.5*4.1*2.62^2 = 14.07 J

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