Question

A body is dropped from a certain height to the ground, when it is halfway down the change in kinetic energy is:
1. More than the change in potential energy
2. Equal to the change in potential energy
3. Less than the change in potential energy
4. None of these

Answer 1

Correct Answer - Option 2 : Equal to the change in potential energy

CONCEPT:

Mechanical energy

• The energy possessed by a body due to its position or state of motion is called mechanical energy.
• Mechanical energy is of two types,
  1. Kinetic energy
  2. Potential energy

Kinetic energy

• The energy possessed by a body due to the virtue of its motion is called kinetic energy.

\(⇒ KE=\frac{1}{2}mv^{2}\)

Where KE = kinetic energy, m = mass and v = velocity

Potential energy

• The energy possessed by a body due to its position is called potential energy.

\(⇒ PE=mgh\)

Where PE = increase in potential energy when an object is taken to height h and g = gravitational acceleration

EXPLANATION:

Let the initial height of the body be h.

When the body is dropped from height h, its velocity is 0 m/sec.

So the initial kinetic energy of the body is, (v₁ = 0 m/sec)

\(⇒ KE_1=\frac{1}{2}mv_1^{2}\)

\(⇒ KE_1=\frac{1}{2}m\times0^{2}\)

⇒ KE₁ = 0      -----(1)

The initial potential energy of the body is, (h₁ = h)

⇒ PE₁ = mgh₁

⇒ PE1 = mgh       -----(2)

At halfway down the height of the body is \(\frac{h}{2}\).

By the third equation of motion,( U = v₁ = 0 m/sec, a = g, and s = \(\frac{h}{2}\))

⇒ V2 = U2 + 2as

\(⇒ v_2^2=v_1^2+2g\frac{h}{2}\)

\(⇒ v_2^2=0^2+2g\frac{h}{2}\)

\(⇒ v_2^2=gh\)        -----(3)

So the final kinetic energy at halfway is given as,

\(⇒ KE_2=\frac{1}{2}mv_2^{2}\)

\(⇒ KE_2=\frac{mgh}{2}\)       -----(4)

And the final potential energy at halfway is given as, (h₂ = \(\frac{h}{2}\))

⇒ PE₂ = mgh₂

\(⇒ PE_2=\frac{mgh}{2}\)       -----(5)

By equation 4 and equation 1 the change in kinetic energy is given as,

⇒ ΔKE = KE₂ - KE₁

\(⇒ \Delta KE=\frac{mgh}{2}-0\)

\(⇒ \Delta KE=\frac{mgh}{2}\)       -----(6)

By equation 5 and equation 2 the change in potential energy is given as,

⇒ ΔPE = PE1 - PE2

\(⇒ \Delta PE=mgh-\frac{mgh}{2}\)

\(⇒ \Delta PE=\frac{mgh}{2}\)       -----(7)

By equation 6 and equation 7,

⇒ ΔKE = ΔPE

Hence, option 2 is correct.