Question

A block of 100 g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given in the question ,
Here mass of the block = 100 g = 0.1 kg 
Velocity v = 5 m/s

Now for Initial Kinetic Energy .
= 1/2 mv² 
= 1/2 (0.100 x 5²)
= 0.5 x 2.5
= 1.25 J.
Now change in Kinetic energy = 0 - 1.25   {Final K.,E = 0]
= -1.25

Therefore diff B/w it's original position & final position =Diameter of tube. 
h= 2 x 10 m
h= 0.20 m

Thus work done by the gravity is 
= mgh
= 01 x 9.8 x 0.2
W = 0.2 J

Here, change in Kinetic energy = work done by gravity - work done by tube.
Thus, Work done by tube = W.D by gravity - Change in Kinetic energy
= 0.2 - (-1.25)
= 0.2 +1.25
= 1.45 J.
Here the movement will always against the force since the W.D is negative.
= -1.45
So the work done by the tube on the block is -1.45 J.


Hope it Helps.

Answer 2

Answer:

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Thanks for asking question