A freely falling body takes 4 s to reach the ground. One second after release, the percentage of its potential energy, that is still retained is

Question

Question

This question was previously asked in

ALP CBT 2 Physics and Maths Previous Paper 3 (Held On: 21 Jan 2019 Shift 3)

Answer 1

CONCEPT:

Potential energy: The energy of an object due to its position is called potential energy. It is denoted by PE.

Mathematically potential energy can be written as

P.E of object = m g h

Where m = mass of an object, g = acceleration due to gravity and h = height

The equation of motion is given by:

S = u t + (1/2) a t2

Where S is displacement, u is initial velocity, a is acceleration and t is time taken.

CALCULATION:

Acceleration due to gravity (g) = 10 m/s²

The body takes (t) 4 sec to reach the ground:

Initial velocity (u) = 0

Acceleration (a) = g = 10 m/s²

Height of the body (H) = ut + (1/2) a t² = 0 + 1/2 × 10 × 4² = 80 m

After 1 sec release:

S = ut + (1/2) a t2 = 0 + 1/2 × 10 × 12 = 5 m

Height (h) = 80 - 5 = 75 m

Potential energy (PE') = m g h = m g × 75 = 75 mg

Initial potential energy (PE) = m g H = 80 mg

% PE = (PE'/PE) × 100 = (75mg/80mg) × 100 = 93.75 %

So option 3 is correct.