A block of mass 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.
Mass of the block, m = 100 g = 0.1 kg
Net displacement in height of the block, h = diameter of the block = 2×10 = 20 cm = 0.2 m
Work done by gravity on the block, W = -mgh
= 0.1 × 9.8 × 0.2
= 0.196 J
Initial kinetic energy of the block = 1/2 × mass of the block × (initial speed of the block)2
= 0.5 × 0.1 × 5 × 5
= 1.25 J
Final kinetic energy of the block = 0 J (Eventually, it stops)
Change in kinetic energy of the block = 0 - 1.25 = -1.25 J
This change in kinetic energy = work done by gravity + work done by the tube
⇒ Work done by the tube = change in kinetic energy - work done by gravity
= -1.25 – 0.196
= -1.45 J