A sled and rider, gliding over horizontal, frictionless ice at 4.0 m/s, have a combined mass of 80.0 kg. The sled then slides over a rough spot in the ice, slowing down to 3.0 m/s. (a) What impulse was delivered to the sled by the friction force from the rough spot? (b) If the sled was in contact with the rough spot is 1.50 s, what was the average force exerted?
Answers
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Answer:
The average force exerted by the friction force from the rough spot was 53.3 N.
Explanation:
From the above question,
They have given :
A sled and rider, gliding over horizontal, frictionless ice at 4.0 m/s, have a combined mass of 80.0 kg.
The sled then slides over a rough spot in the ice, slowing down to 3.0 m/s.
(a) Impulse is defined as the change in momentum.
In this case, the initial momentum of the sled and rider is 80.0 kg * 4.0 m/s = 320.0 kg m/s. After sliding over the rough spot, the final momentum is 80.0 kg * 3.0 m/s = 240.0 kg m/s. Therefore, the change in momentum is 320.0 kg m/s - 240.0 kg m/s = 80.0 kg m/s. This means that the impulse delivered to the sled by the friction force is 80.0 kg m/s.
(b) The average force exerted is equal to the impulse divided by the time it was exerted for:
Average force = Impulse / Time
= 80.0 kg m/s / 1.50 s
= 53.3 N
Therefore, the average force exerted by the friction force from the rough spot was 53.3 N.
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