A scooter starting from rest moves with a constant acceleration for a time Δt1, then with a constant velocity for the next Δt2, and finally with a constant deceleration for the next Δt3, to come to rest. A 500 N man sitting on the scooter behind the driver manages to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is
(a) 500 N throughout the journey
(b) less than 500 N throughout the journey
(c) more than 500 N throughout the journey
(d) > 500 N for the time Δt1, and Δt3 and 500N for Δt2.
Answers
Answer ⇒ Option (d). > 500 N for the time Δt₁, and Δt₃ and 500 N for Δt₂.
Explanation ⇒ During the time Δt₂, the Scooter moves with the constant velocity, therefore there will be no friction. The only force acting on it will be the force of the Normal Reaction which will be equal to the 500 N.
But during the time impact of the Δt₁ and Δt₃ there is the acceleration and deceleration respectively. Thus, there is acting the Force of the Friction along with the Normal Reaction.
Now, the Resultant of the Normal Reaction and the Friction will give a force of the magnitude greater than 500 N,
because, F (resultant) = √[(500)² + (fric.)²]
Hope it helps.
Answer:
Answer ⇒ Option (d). > 500 N for the time Δt₁, and Δt₃ and 500 N for Δt₂.
Explanation ⇒ During the time Δt₂, the Scooter moves with the constant velocity, therefore there will be no friction. The only force acting on it will be the force of the Normal Reaction which will be equal to the 500 N.
But during the time impact of the Δt₁ and Δt₃ there is the acceleration and deceleration respectively. Thus, there is acting the Force of the Friction along with the Normal Reaction.
Now, the Resultant of the Normal Reaction and the Friction will give a force of the magnitude greater than 500 N,
because, F (resultant) = √[(500)² + (fric.)²]
Hope it helps.