Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
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Below is the solution:
9.43 m/s
<span>F = -kx </span>
<span>k = 400 N/m </span>
<span>PE= 0.5kx^2 = 0.5mv^2 </span>
<span>solve, v=9.43 m/s</span>
Answer:
false
Explanation:
I had the same answer for this question on my schoolwork
:) HOPE THIS HELPS
They're equal.
Ignoring air resistance, any object you're not holding onto has the same, constant acceleration ... 9.8 m/s^2 down ... from the moment it leaves your hand until the moment it hits the ground.
It doesn't matter how heavy the object is, and it doesn't matter whether you throw it up, down, or sideways, or just let go and let it drop.
C most of the Earths weather occurs in this layer.
