Question

A 5kg ball is dropped from a height of 10 m .
a. Find the initial potential energy of the ball.
b. Calculate the velocity before it reaches the ground .
c. Find the kinetic energy just before it reaches the ground .

Answer 1

Answer:

Here mass of the ball, m=5kg

height of the ball, h=10m

(a) initial potential energy of the ball,

Ep=mgh=(5kg)(10m/s^2)(10m)=500J

(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into

its kinetic energy

Ek=500J.

It υ is the velocity attained by the ball before reaching the ground,

Ek=12mυ^2

14.14m/s

Answer 2

A 5kg ball is dropped from a height of 10m.

1. Find the initial potential energy of the ball
2. the velocity before it reaches the ground
3. the kinetic energy just before it reaches the ground

1. initial potential energy , U = mgh

m = mass of ball = 5 kg

g = acceleration due to gravity = 10 m/s²

h = height of ball = 10 m

∴ U = 5 kg × 10 m/s² × 10 m = 500 J

Therefore the initial potential energy of the ball is 500J.

2. initial velocity of ball, u = 0

distance covered by ball before it reaches the ground, s = 10 m

using formula, v² = u² + 2as

⇒v² = 0² + 2(10)(10)

⇒v² = 200

⇒v = 10√2 m/s

Therefore the velocity of the ball before it reaches the ground is 10√2 m/s.

3. kinetic energy just before it reaches the ground, K = 1/2 mv²

= 1/2 × 5kg × (10√2 m/s)²

= 1/2 × 5kg × 200 m²/s²

= 500 J

Therefore the kinetic energy just before it reaches the ground is 500 J.