Question

A 5 kg ball is dropped from a height of 10 m. (a)find initial potential energy of the ball. (b)calculate the velocity before it reaches the ground​

Answer 1

Answer:

(a) Initial Potential Energy = 500 J

(b) the velocity before the ball reaches the ground = 14.14 m/s

Explanation:

(a) The potential energy is given by,

PE = mgh

where

m denotes mass of the object

g denotes the acceleration due to gravity (10 m/s²)

h denotes the height from the ground

Substitute,

PE = 5 × 10 × 10

PE = 500 J

Therefore, initial potential energy of the ball is 500 J

(b) After the ball has reached the ground, potential energy is 0 J. As the total energy is constant, the initial potential energy is converted to kinetic energy.

Kinetic energy = 500 J

KE = ½ mv²

where

v denotes the velocity

Substitute,

500 = ½ × 5 × v²

v² = 200

v = √200

v = 10√2

v = 14.14 m/s

Therefore, the velocity before it reaches the ground is 14.14 m/s

Answer 2

Given

• Mass (m) of ball =  5 Kg
• Height (h) = 10 m

To find

1. Initial potential energy
2. Velocity before the ball reaches ground

Solution

We know, potential energy can be found with the formula :-

• P.E = mgh

Where :-

• m - mass (5)
• g - acceleration due to gravity (9.8)
• h - height (10)

Substituting the values, we get :-

• P.E = mgh
• P.E = (5)(9.8)(10)
• P.E = 490

(a) Hence, the potential energy is 490 J

After the ball reaches ground, there will be kinetic energy acting on the body, potential energy will be 0 J.

We know, kinetic energy can be found with the formula :-

• K.E = 1/2mv²

Where :-

• m - mass (5)
• K.E = (490)
• v - velocity

Substituting the value, we get :-

• 490 = 1/2(5)(v)²
• v² = 196
• v = √196
• v = 14

(b) Hence, the velocity is 14 m/s