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when her speed reaches 1.1 m/s, she pulls in such a way that she and the platform go up at a constant speed. what force is she exerting on the rope now? (ignore the mass of the rope.)

1 Answer

6 votes

The force exerted by the student on the rope is equal to the tension in the rope, which is equal to the weight of the student and the platform. The tension can be calculated by multiplying the combined mass of the student and the platform by the acceleration due to gravity.

In this scenario, the student exerts a force on the rope to go up with the platform at a constant speed. The force she exerts on the rope is equal to the tension in the rope. Since the student and the platform are going up at a constant speed, the net force on them is zero. Therefore, the tension in the rope is equal to the weight of the student and the platform.

In the equation F = m times g, where F is the force, m is the mass, and g is the acceleration due to gravity, the mass of the student and the platform can be multiplied by the acceleration due to gravity to determine the tension in the rope.

For example, if the combined mass of the student and the platform is 50 kg and the acceleration due to gravity is 9.8 m/s^2, then the tension in the rope would be 9.8 * 50 = 490 N (Newtons).

answered
User Steve Bauman
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8.1k points
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