Anne releases a stone from a height of 2 meters. She measures the kinetic energy of the stone at 9.8 joules at the exact point i
2 answers:
The correct answer to the question is A) 0.5 Kg.
CALCULATION:
Before going to answer this question, first we have to understand the law of conservation of energy.
As per law of conservation of energy, the energy can neither be created nor be destroyed. It can only change from one form to another form and the total energy of universe is always constant.
As per the question, the stone was initially at a height of 2m from the ground.The energy possessed by the stone at this height is gravitational potential energy.
When the stone is released, the stone moves downward under gravity. It gains kinetic energy. When it just hits the ground, whole of its potential energy must have been converted into kinetic energy.
Hence, P.E at height 2m = K.E of the stone just before hitting the ground.
The potential energy of a body at a height h from the surface of earth is calculated as-
Potential energy P.E = mgh
Here, g is known as the acceleration due to gravity and its value is .
It is given that the height of the stone h = 2m.
The kinetic energy of the stone just before reaching ground K.E = 9.8 J.
We know- P.E of stone at height 2m = K.E of stone when it hits the ground.
⇒ mgh = 9.8 Joules
⇒ m =
=
= [ans]
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Answer:
i) The period of the particle is 0.3 seconds
ii) The angular velocity is approximately 20.94 rad/s
iii) The linear velocity is approximately 1.047 m/s
iv) The centripetal acceleration is approximately 6.98 m/s²
Explanation:
The given parameters are;
The number of revolution of the particle, n = 800 revolution
The time it takes the particle to make 800 revolutions = 4 minutes
The dimension of the circle = 5 cm = 0.05 m
Given that the dimension of the circle is the radius of the circle, we have;
i) The period of the particle, T = The time to complete one revolution
T = 1/(The number of revolutions per second)
∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s
The period, T = 3/10 seconds = 0.3 seconds
ii) The angular velocity, ω = Angle covered/(Time)
800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes
∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3
The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s
iii) The linear velocity, v = r × ω
∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s
iv) The centripetal acceleration, = v²/r
∴ The centripetal acceleration, = (π/3)²/(0.05) = 20·π/9
The centripetal acceleration, = 20·π/9 m/s² ≈ 6.98 m/s²