Question

An object of mass ‘m’ is allowed to fall freely from point A as shown in the figure. Calculate the total mechanical energy of the object at:

1. Point A
2. Point B
3. Point C

State the law which is verified by your calculations in parts (i), (ii) and (iii).

Answer 1

(i) Since at point A, height is x + y and velocity = 0

Total energy at A |= P. E. at A + K. E. at A

= `mg (x + y) + 1/2 xx m xx 0^2`

= mg (x + y)

(ii) At point B, height is y and velocity = v₁

Total energy at B = P. E. at B + K. E. at B

= `mg  y + 1/2 mv_1^2`

But `v_1^2` = u² + 2 gh

`v_1^2` = 2 gx  ....[∵ u = 0 during free fall]

∴ Total energy at B = `mg  y + 1/2 m xx 2  gx`

= mg (x + y)

(iii) At point C, height = 0 and velocity = v₂

Total energy at C = P. E. at C + K. E. at C

= `0 + 1/2 mv_2^2`

But `v_2^2` = u² + 2 gh

`v_2^2` = 0 + 2g (x + y)

`v_2^2` = 2g (x + y)

∴ Total energy at C = `1/2 m xx 2g (x + y)`

= mg (x + y)

Since the total energy at points A, B, and C is exactly the same, i.e., mg (x + y), the law of conservation of energy is successfully verified.