Activity 11.15 - An object of mass 20 kg is dropped from a height of 4 m

Last updated at March 16, 2023 by Teachoo

An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case. Take g = 10 m/s
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Activity 11.15 An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case. Take g = 10 m/s2
Mass of the object = m = 20 kg
Acceleration due to gravity = g = 10 m/s2
At Height = 4 m
Since the is just dropped,
Velocity of the ball = v = 0 m/s
Potential Energy
Potential Energy = m × g × h
Ep = 20 × 10 × 4
Ep = 800 J
Kinetic Energy
Kinetic Energy = 1/2 mv2
Ek = 1/2 × 20 × 02
Ek = 0 J
Total energy = Ep + Ek
= 800 + 0
= 800 J
At Height = 3 m
To find Potential Energy, we need height
To find Kinetic Energy, we need velocity
Finding Velocity
Initial velocity = u = 0 m/s
Acceleration = a = g = 10 m/s
Distance travelled = s = 4 − 3
= 1 m
We know u, a and s
We can find v using 3rd equation of motion
v2 − u2 = 2as
(Velocity at highest point is 0)
v2 − u2 = 2as
v2 − 0 = 2 × 10 × 1
v2 = 20
Finding Potential and Kinetic Energy
Potential Energy
Potential Energy = m × g × h
Ep = 20 × 10 × 3
Ep = 600 J
Kinetic Energy
Kinetic Energy = 1/2 mv2
Ek = 1/2 × 20 × 20
Ek = 200 J
Total energy = Ep + Ek
= 600 + 200
= 800 J
At Height = 2 m
To find Potential Energy, we need height
To find Kinetic Energy, we need velocity
Finding Velocity
Initial velocity = u = 0 m/s
Acceleration = a = g = 10 m/s
Distance travelled = s = 4 − 2
= 2 m
We know u, a and s
We can find v using 3rd equation of motion
v2 − u2 = 2as
(Velocity at highest point is 0)
v2 − u2 = 2as
v2 − 0 = 2 × 10 × 2
v2 = 40
Finding Potential and Kinetic Energy
Potential Energy
Potential Energy = m × g × h
Ep = 20 × 10 × 2
Ep = 400 J
Kinetic Energy
Kinetic Energy = 1/2 mv2
Ek = 1/2 × 20 × 40
Ek = 400 J
Total energy = Ep + Ek
= 400 + 400
= 800 J
At Height = 1 m
To find Potential Energy, we need height
To find Kinetic Energy, we need velocity
Finding Velocity
Initial velocity = u = 0 m/s
Acceleration = a = g = 10 m/s
Distance travelled = s = 4 − 1
= 3 m
We know u, a and s
We can find v using 3rd equation of motion
v2 − u2 = 2as
(Velocity at highest point is 0)
v2 − u2 = 2as
v2 − 0 = 2 × 10 × 3
v2 = 60
Finding Potential and Kinetic Energy
Potential Energy
Potential Energy = m × g × h
Ep = 20 × 10 × 1
Ep = 200 J
Kinetic Energy
Kinetic Energy = 1/2 mv2
Ek = 1/2 × 20 × 60
Ek = 600 J
Total energy = Ep + Ek
= 200 + 600
= 800 J
Just above the ground (h ≈ 0 m)
To find Potential Energy, we need height
To find Kinetic Energy, we need velocity
Finding Velocity
Initial velocity = u = 0 m/s
Acceleration = a = g = 10 m/s
Distance travelled = s = 4 − 0
= 4 m
We know u, a and s
We can find v using 3rd equation of motion
v2 − u2 = 2as
(Velocity at highest point is 0)
v2 − u2 = 2as
v2 − 0 = 2 × 10 × 4
v2 = 80
Finding Potential and Kinetic Energy
Potential Energy
Potential Energy = m × g × h
Ep = 20 × 10 × 0
Ep = 0 J
Kinetic Energy
Kinetic Energy = 1/2 mv2
Ek = 1/2 × 20 × 80
Ek = 800 J
Total energy = Ep + Ek
= 0 + 800
= 800 J
Thus, our table looks like

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo

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