A mass m 1 = 5.2 kg rests on a frictionless table and connected by a massless st

Question

A mass m1 = 5.2 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 5.2 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.85 m.

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

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

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

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

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

6)The work done by tension on only m2 is:

positive

zero

negative

7)What is the NET work done on m2?

Explanation / Answer

1.
m1 will move a distance of d horizontally.
Since force due togravity on m1 is perpendicular to its displacement.
work done on m1 = 0

work done on m2= F*d
= m2*g*d
=5.2*9.8*0.85
= 43.32 J

2)
Normal force on m1 is perpendicular to displacement.
so, work done = 0

3)
net work done on both block by external force = 43.32 J
Since gravity is only external force. Tension is internal force
Speed on both block will be same.
consider m2,
work done= increase in kinetic energy
43.32 = 0.5*m2*v^2 + 0.5*m1*v^2
43.32 = 5.2*v^2
v = 2.89 m/s
for both block

4)
work done by tension = increase in kinetic enrgy of m1
= 0.5*m1*v^2
= 0.5*5.2*2.89^2
=21.66 J

5)
work done = T*d
21.66 = T*0.85
T= 25.5 N

6)
on m2, tension is opposite in direction of motion.
so, workdone is negative

7)

Net work done on m2 = 0

work done by gravity and tensions cancels each other

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