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A block is initially at rest at the top of a 8.0 m high hill. The block slides down to the bottom of the hill and the back up a second
Question
A block is initially at rest at the top of a 8.0 m high hill. The block slides down to the
bottom of the hill and the back up a second hill with a height of 5.4 m. What is the
speed of the block at the top of the second hill. Assume no frictional forces.
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Physics
5 years
2021-08-15T03:58:17+00:00
2021-08-15T03:58:17+00:00 1 Answers
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Answers ( )
Answer:
The speed of the block at the top of the second hill = 7.14 m/s
Explanation:
According to the principle of conservation of energy, the potential energy lost by the block due to decrease in elevation is equal to the kinetic energy gained by the block due to motion
The initial potential energy of the block is found from the equation of potential energy as follows;
Where the condition of no friction is applied
PE = Mass, m× Acceleration due to gravity, g × Height, h
Where:
m = The mass of the block
g = 9.81 m/s²
h is given as 8.0 m
The kinetic energy, KE, gained by the sliding block = 1/2 × mass of the block, m × (Velocity of the block, v)²
That is KE = 1/2×m×v²
At the bottom of the hill the potential energy is completely converted into potential energy which gives;
m × 9.81 × 8 = 1/2×m×v²
KE = 1/2×m×v² = m×78.48
9.81 × 8 = 1/2×v² (By cancelling the like term on both sides of the above equation)
v² = 2 × 9.81 × 8 = 156.96
v = √156.96 = 12.53 m/s
At the height h = 5.4 m. of the second hill, we have;
Potential energy gained = Kinetic energy lost
Kinetic energy remaining = Initial kinetic energy – Kinetic energy lost
Kinetic energy remaining = Initial kinetic energy – Potential energy gained
Kinetic energy remaining = m×78.48 – m × 9.81 × 5.4 = m×(78.48 – 52.974)
1/2×m×v₂² = m×25.506
Where:
v₂ = The speed of the block at the top of the second hill
v₂² = 2×25.506 = 51.012
v₂ = 7.14 m/s
The speed of the block at the top of the second hill = 7.14 m/s.