Question

A man tries to lift a mass 200kg with a force 100N.

1. Is he doing work? Explain.

2. If it is lifted to 2m in 10s, find the power.

3. Show that total mechanical energy is conserved for a freely falling body.

Answer 1

1. No. Force required to lift the body is 2000N (w = mg = 200 × 10). 

But the applied force is 100N. Hence there is no displacement due to this applied force.

2. Power P = \(\frac{w}{t} = \frac{mgh}{t} = \frac{200×10×2}{10}\) = 400 watt.

3.

Consider a body of mass ‘m’ at a height h from the ground.

Total energy at the point A

Potential energy at A,

PE = mgh

Kinetic energy, KE = \(\frac{1}{2}\) mv² = 0

(since the body at rest, v = 0).

∴ Total mechanical energy = PE + KE

= mgh + 0 = mgh

Total energy at the point B

The body travels a distance x when it reaches B. The velocity at B, can be found using the formula.

v² = u² + 2as

v² = 0 + 2 gx

∴ KE at B, = \(\frac{1}{2}\) mv²

= \(\frac{1}{2}\)m²gx

= mgx

P.E. at B, = mg(h – x)

Total mechanical energy = PE + KE

= mg(h – x) + mgx

= mgh

Total energy at C

Velocity at C can be found using the formula

v² = u² + 2as

v² = 0 + 2gh

∴ KE at C, = \(\frac{1}{2}\)mv²

= \(\frac{1}{2}\)m²gh

= mgh

P.E. at C = 0

Total energy = PE + KE = 0 + mgh = mgh.