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)
K.E₂ = mg(h - 2R)
The kinetic energy of the car is expressed as K.E = mg(h - 2R). The energy of the object by the integrity of its movement is understood as kinetic energy.
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of velocity.
The following data is observed from the figure;
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0
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)
According to the law of conservation of energy, total energy is defined as the sum of kinetic energy and potential energy.
Total energy = kinetic energy+potential energy
On applying the law of conservation of energy for top and bottom of the loop;
K.E₁ + P.E₁ = K.E₂ + P.E₂
0 + mgh = K.E₂ + mg(2R)
K.E₂ = mg(h - 2R)
Hence the kinetic energy of the car is expressed as K.E = mg(h - 2R).
To learn more about kinetic energy refer to the link;
brainly.com/question/999862
