1. A 1-kg body falls freely from rest, from a height of 80 m. Acceleration due to gravity is 10 m/s2. What is the kinetic energy when the body hits the ground.
Known :
Mass (m) = 1 kg
Height (h) = 80 m
Acceleration due to gravity (g) = 10 m/s2
Wanted: kinetic energy when the body hits the ground
Solution :
The initial mechanical energy (MEo) = gravitational potential energy (PE)
MEo = PE = m g h = (1)(10)(80) = 800 Joule
The final mechanical energy (MEt) = kinetic energy (KE)
The principle of conservation of mechanical energy :
MEo = MEt
PE = KE
800 = KE
The final kinetic energy is 800 Joule.
2. A 4-kg body free fall from rest, from a height of 10 m. Acceleration due to gravity is 10 m s–2. What is the kinetic energy and the velocity at 5 meters above the ground.
Known :
The change in height (h) = 10 – 5 = 5 meters
Mass (m) = 4 kg
Acceleration due to gravity (g) = 10 m/s2
Wanted: Kinetic energy at 5 meters above the ground and the velocity at 5 meters above the ground
Solution :
(a) Kinetic energy at 5 meters above the ground
The initial mechanical energy (MEo) = the gravitational potential energy (PE)
MEo = PE = m g h = (4)(10)(5) = 200 Joule
The final mechanical energy (EMt) = kinetic energy (EK)
MEt = KE
The principle of conservation of mechanical energy states that the initial mechanical energy = the final mechanical energy.
MEo = MEt
200 = KE
Kinetic energy at 5 meters above the ground is 200 Joule.
(b) velocity at 5 meters above the ground
The initial mechanical energy (MEo) = the final mechanical energy (MEt)
PE = KE
200 = ½ m v2
2(200) / 4 = v2
100 = v2
v = √100
v = 10 m/s
Body’s velocity at 5 meters above the ground is 10 m/s.
3. A mango falls freely from rest, from a height of 2 meters. Acceleration due to gravity is 10 m s–2. Determine mango’s velocity when hits the ground.
Known :
Height (h) = 2 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted : mango’s velocity when hits the ground.
Solution :
The initial mechanical energy (MEo) = the gravitational potential energy (PE)
ME = PE = m g h = m (10)(2) = 20 m
The final mechanical energy (MEt) = the kinetic energy (KE)
MEt = KE = ½ m v2
Principle of conservation of mechanical energy states that the initial mechanical energy = the final mechanical energy.
MEo = MEt
20 m = ½ m v2
20 = ½ v2
2(20) = v2
40 = v2
v = √40 = √(4)(10) = 2√10 m/s
[wpdm_package id=’1166′]
- Work done by force problems and solutions
- Work-kinetic energy problems and solutions
- Work-mechanical energy principle problems and solutions
- Gravitational potential energy problems and solutions
- The potential energy of elastic spring problems and solutions
- Power problems and solutions
- Application of conservation of mechanical energy for free fall motion
- Application of conservation of mechanical energy for up and down motion in free fall motion
- Application of conservation of mechanical energy for motion on a curved surface
- Application of conservation of mechanical energy for motion on an inclined plane
- Application of conservation of mechanical energy for projectile motion