Two people with a mass of 70 Kg wearing inline skates and are holding opposite sides of a 15 m rope. One person...... pulls forward on the rope by moving hand over hand and gradually reeling in more...
Two people with a mass of 70 Kg wearing inline skates and are holding opposite sides of a 15 m rope. One person......
pulls forward on the rope by moving hand over hand and gradually reeling in more of the rope. In doing so, he exerts a force of 35 N(backwards) on the rope. This causes him to accelerates towards the other person. Assuming that the friction acting on the skates is negligible, how long will it take for them to meet? Explain
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Here first person pulls the rope by exerting a backward force on it. Due to that he will move forward because of the reaction of the force on rope will act on the person. On the other side due to the tension of the rope because of the pull the second person will move towards the first person.
How ever only first person reeling the rope. The second one only holds it up. So it is clear that when they meet they have to travel 15m together. Since the forces are equal each will travel 15/2 = 7.5m
Using `F = ma` to second persons direction
`35 = 70xxa`
`a = 0.5m/s`
`S = Ut+1/2xxaxxt^2` second persons direction
`7.5 = 0+1/2xx0.5xxt^2`
`t = 2.74`
So it will take 2.74 seconds to meet each other.