Question

1.
(d) kinetic energy of the body at the end of the trip?
A block of mass 200 g is moving with a speed of 4 m/s at the highest point in a closed circular tube of
radius 10 cm kept fixed 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. (g = 10 m/s)
horizontal table if the friction coefficient between block​

Answer 1

Answer:

mass-200g=0.2Kg

speed 4m/s

displacement 10cm=0.01m

a =10m/s^2

W.D= Force×Displacement

Force=Ma

=0.2×10m/s^2

2newton

Work done = 2×0.01=1/50=0.02 Joules

Answer 2

Answer:

Mass = 200 g = 0.2 kg

Speed = 4 m/s

Displacement = 10 cm = 0.01 m

Acceleration = 10 m/s²

Work Done = Force × Displacement

W.D. = F × S

Force = Mass × Acceleration

F = ma

0.2 × 10 m/s²

= 2 Newton or 2 N

Work Done = 2 × 0.01 = 0.02 Joule or 0.02 J.

I hope it helps you.