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1. A person pulls a heavy load of mass 37 kg up the side of a building by using a frictionless pulley. The load travels up a dis- tance of 7.5 m. Take the load to move with constant velocity and ignore any acceleration at the beginning or end of the move. (a) How much work is done on the load by gravity? (b) By the tension of the rope? (c) By the person?

User CephBirk
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Final answer:

a. The work done by gravity on the load is -3427.5 joules.

b and c. The work done by the tension in the rope and by the person is 3427.5 joules each, assuming no other forces are at work.

Step-by-step explanation:

The question pertains to work done by various forces when a person pulls a heavy load using a frictionless pulley.

(a) The work done on the load by gravity is calculated by the formula Work = Force x Distance x cos(θ).

Since the direction of the gravitational force is directly downwards and the displacement is upwards, θ = 180 degrees, and cos(θ) = -1 (since cos(180) = -1).

The force due to gravity (weight) is mass x gravitational acceleration (W = mg).

So, Work by gravity = mg x distance x (-1).

Here, m = 37 kg, g = 9.8 m/s² (standard gravity), and distance = 7.5 m. Thus, Work by gravity = 37 kg x 9.8 m/s² x 7.5 m x (-1) = -3427.5 J.

The work by gravity is negative because gravity is acting in the opposite direction of the displacement.

(b) Since the tension of the rope acts upwards and the load is moved upwards at a constant velocity, the tension in the rope is equal to the weight of the load. Therefore, the work done by the tension will be equal to the magnitude of the work done by gravity but it will be positive because the tension force and the displacement are in the same direction. Work by tension = 3427.5 J.

(c) The work done by the person will be equal to the work done by the tension since the person is applying this force through the rope, assuming no other forces like friction are at play. So, Work by person = 3427.5 J.

User Pomkine
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