A 3kg box has 45J of gravitational potential energy, how high is it off the ground?
1 answer:
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Answer:
t = 2.01 s
Vf = 19.7 m/s
Explanation:
It's know through the International System that the earth's gravity is 9.8 m/s², then we have;
Data:
- Height (h) = 20 m
- Gravity (g) = 9.8 m/s²
- Time (t) = ?
- Final Velocity (Vf) = ?
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Time
Use formula:
Replace:
Everything inside the root is solved first. So, we solve the multiplication of the numerator:
It divides:
The square root is performed:
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Final Velocity
use formula:
- Vf = g * t
Replace:
- Vf = 9.8 m/s² * 2.01 s
Multiply:
- Vf = 19.7 m/s
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How long does it take to reach the ground?
Takes time to reach the ground in <u>2.01 seconds.</u>
How fast does it hit the ground?
Hits the ground with a speed of <u>19.7 meters per seconds.</u>
Answer:
(a) The maximum height achieved by the first ball, m₁ is 0.11 m
(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m
Explanation:
Given;
mass of the first ball, m₁ = 1 kg
mass of the second ball, m₂ = 2 kg
The velocity of the first when released from a height of 1 m before collision;
u₁² = u₀² + 2gh
u₀ = 0, since it was released from rest
u₁² = 2gh
u₁² = 2 x 9.8 x 1
u₁² = 19.6
u₁ = √19.6
u₁ = 4.427 m/s
The velocity of the second ball before collision, u₂ = 0
Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
v₁ is the final velocity of the first ball after an elastic collision
v₂ is the final velocity of the second ball after an elastic collision
m₁u₁ + m₂(0) = m₁v₁ + m₂v₂
m₁u₁ = m₁v₁ + m₂v₂
1 x 4.427 = v₁ + 2v₂
v₁ + 2v₂ = 4.427
v₁ = 4.427 - 2v₂ ----- equation (1)
one directional velocity;
u₁ + v₁ = u₂ + v₂
u₂ = 0
u₁ + v₁ = v₂
v₁ = v₂ - u₁
v₁ = v₂ - 4.427 ------ equation (2)
Substitute v₁ into equation (1)
v₂ - 4.427 = 4.427 - 2v₂
3v₂ = 4.427 + 4.427
3v₂ = 8.854
v₂ = 8.854 / 3
v₂ = 2.95 m/s (→ forward direction)
v₁ = v₂ - 4.427
v₁ = 2.95 - 4.427
v₁ = - 1.477 m/s
v₁ = 1.477 m/s ( ← backward direction)
Apply the law of conservation of mechanical energy
(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)
(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)