a stone of mass 0.2 kg is thrown vertically up with a speed of 19.6 m/a. calculate its potential energy and kinetic energy after 1s
Answers
Answer:
Potential energy = 28.812 joules
Kinetic energy = 9.604 joules
Explanation:
Given:
Mass of the stone = m = 0.2 kg
Velocity with which the stone is thrown = u = 19.6 m/s
Time = 1 second
Gravitational force = g = 9.8 m/s²
To find:
- Potential energy
- Kinetic energy
Potential energy:
Potential energy = m×g×h
So, we need to find the height here,
Using second equation of motion which says :
s=19.6×1
s=19.6-4.9
s=14.7 metres
The stone will travel a distance of 14.7 metres
Potential energy = 0.2×9.8×14.7
Potential energy = 28.812 joules
Kinetic energy:
Here we need to find the value of final velocity (v)
Using third equation of motion:
v²-u²=2as
v²-19.6²=2×-9.8×14.7
v²-384.16= -288.12
v² = -288.12+384.16
v² = 96.04
v = 9.8 m/s
Kinetic energy =
Kinetic energy = 0.5×0.2×9.8×9.8
Kinetic energy = 0.1×96.04
Kinetic energy = 9.604 joules
Given :
▪ Mass of stone = 0.2kg
▪ Initial velocity = 19.6mps
To Find :
▪ Potential energy and kinetic energy of ball after 1s
Solution :
→ For potential energy, we have to find out height attained by stone after 1s.
→ For kinetic energy, we have to find out final velocity of stone after 1s.
→ Since, acceleration due to gravity is constant throughout the motion, we can easily apply equations of kinematics to solve this type of questions.
✴ Height attained by ball after 1s :
✏ H = ut - (1/2)gt^2
Negative sign shows opposite direction
✏ H = (19.6×1) - (0.5×9.8×1)
✏ H = 19.6 - 4.9
✏ H = 14.7m
✴ Potential energy of ball :
✒ U = mgH
✒ U = 0.2×9.8×14.7
✒ U = 28.812J
✴ Final velocity of ball after 1s :
✏ v^2 - u^2 = -2gH
✏ v^2 - (19.6)^2 = -2(9.8)(14.7)
✏ v^2 - 384.16 = -288.12
✏ v^2 = -288.12 + 384.16
✏ v^2 = 96.04
✏ v = 9.8mps
✴ Kinetic energy of ball :
✒ K = (1/2)mv^2
✒ K = 0.5×0.2×96.04