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BabaBlast [244]
3 years ago
14

two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

orward on the rope by moving hand over hand and gradually reeling in more of the rope. In doing so, he exerts a force of 35N [backwards] on the rope. This causes him to accelerate toward the other person. Assuming that the friction acting on the skaters is negligible, how long will it take for them to meet? Explain your reasoning.
Physics
1 answer:
Zina [86]3 years ago
5 0

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

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(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

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<h3>Free body diagram</h3>

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<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

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Learn more about net force here: brainly.com/question/14361879

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