A ball of mass 0.20 kg is kept at rest at a height 5 m. (acceleration due to gravity= 10 m/s²) Find its potential energy. Now the ball is allowed to fall freely. Find out kinetic energy and the speed of the ball just before it hits the ground. Choose the option with correct answers. *
Answers
Answer:
Mass = 0.20kg
= 200g
Height = 5m
Acc. due to gravity = 10m/s^2
Potential energy = mgh
= 200*10*5
= 10000J
= 10KW J
Speed = v^2-u^2=2as
= v^2-(0)^2=2*10*5
= v^2=100
= v =√100
= v = 10m/s
Kinetic energy = 1/2mv^2
= 1/2*200*10*10
= 10000J
= 10KW J
Answer :
- The gravitational potential energy of the ball of mass 0.2 kg kept at a height of 5 m, P.E = 10 J
- Velocity of the body just before reaching the ground, v = 10 m/s
- Kinetic energy possessed by the ball just before reaching the ground, K.E. = 10 J
Explanation :
Given :
- Mass of the ball, m = 0.2 kg
- Height at which the ball is kept, h = 5 m
- Acceleration due to gravity, g = 10 m/s²
To find :
- Potential energy of the ball when kept at the height of 5 m, P.E. = ?
- Velocity of the body just before reaching the ground, v = ?
- Kinetic energy possessed by the body just before reaching the ground, K.E. = ?
Knowledge required :
⠀⠀⠀● By definition, potential energy possessed by the body due to it's state of rest.
Formula for potential energy :
⠀⠀⠀⠀⠀⠀⠀⠀⠀P.E. = mgh
Where,
- P.E. = Potential energy
- m = Mass
- g = Acceleration due to gravity
- h = Height
⠀⠀⠀● By definition, kinetic energy possessed by the body due to it's state of motion.
Formula for potential energy :
⠀⠀⠀⠀⠀⠀⠀⠀⠀K.E. = ½mv²
Where,
- K.E. = Kinetic energy
- m = Mass
- v = Velocity of the body
⠀⠀⠀● Third Equation of Motion : (Under gravity)
⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² + 2gh
Where,
- v = Final velocity
- u = Initial velocity
- g = Acceleration due to gravity
- h = Height
Solution :
⠀⠀⠀⠀⠀⠀⠀Potential energy of the ball :
By using the equation for potential energy and substituting the values in it, we get :
⠀⠀⠀=> P.E. = mgh
⠀⠀⠀=> P.E. = 0.2 × 10 × 5
⠀⠀⠀=> P.E. = 10
⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ P.E. = 10 J
⠀⠀⠀⠀⠀⠀⠀Velocity of the ball :
By using the third equation of motion and substituting the values in it, we get :
⠀⠀⠀=> v² = u² + 2gh
⠀⠀⠀=> v² = 0² + 2gh
[Note : The Initial velocity of the ball will be zero, since the body is moving with the same velocity as before reaching the ground]
⠀⠀⠀=> v² = 2gh
⠀⠀⠀=> v² = 2 × 10 × 5
⠀⠀⠀=> v² = 100
⠀⠀⠀=> v = √100
⠀⠀⠀=> v = 10
⠀⠀⠀⠀⠀⠀⠀⠀∴ v = 10 m/s
⠀⠀⠀⠀⠀⠀⠀Kinetic energy of the ball :
By using the equation for potential energy and substituting the values in it, we get :
⠀⠀⠀=> K.E. = ½mv²
⠀⠀⠀=> K.E. = ½ × 0.2 × 10²
⠀⠀⠀=> K.E. = ½ × 0.2 × 100
⠀⠀⠀=> K.E. = 10
⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ K.E. = 10 J
Therefore,
- Potential energy of the ball when kept at the height of 5 m, P.E. = 10 J
- Velocity of the body just before reaching the ground, v = 10 m/s
- Kinetic energy possessed by the body just before reaching the ground, K.E. = 10 J