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>
Explanation:
Wooho Physics my favorite subject! And my 100th answer :)
We will use the formula

- Where v is Final velocity
- And u is Initial velocity.
- t is time taken.
In this question :-
- u is 4 m/s
- v is 6 m/s
- t is 5 seconds
Applying the formula:-


So, Acceleration is 0.4 m/s² is the answer.
Hope it helps :)
Choice 'D' describes speed in the metric units 'meter' and 'second'.
With no mention of direction, it can't be called 'velocity'.
Answer:
hello your question is incomplete attached below is the complete question
Answer : attached below
Explanation:
A) finding the particular solution at t = 0
attached below is the detailed solution of the particular solution knowing that t = 0