A conveyor belt ramp of length 7m is inclined at 20° with respect to the horizontal and is used to move objects from one location to another location. When an object is placed on the conveyor belt, the object travels up the incline with a constant speed of 1.5 ms. A motor is used to keep the conveyor belt moving. The mass of the objects that are placed on the conveyor belt are varied. Suppose the motor breaks and a person must push an object of mass 30kg from the bottom of the incline to the top of the incline. The person can exert a force of 150N on the object parallel to the incline. What is the approximate amount of work that the person does on the object to move it to the top of the incline? Assume that all frictional forces are negligible. 0J

Question

Physics

Kaylee

16 September, 01:30

A conveyor belt ramp of length 7m is inclined at 20° with respect to the horizontal and is used to move objects from one location to another location. When an object is placed on the conveyor belt, the object travels up the incline with a constant speed of 1.5 ms. A motor is used to keep the conveyor belt moving. The mass of the objects that are placed on the conveyor belt are varied. Suppose the motor breaks and a person must push an object of mass 30kg from the bottom of the incline to the top of the incline. The person can exert a force of 150N on the object parallel to the incline. What is the approximate amount of work that the person does on the object to move it to the top of the incline? Assume that all frictional forces are negligible. 0J

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Answers (1)

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Fabian Savage · Answer 1

The work done by the person is 1050 J.

Explanation:

Hi there!

The work done by a force applied parallel to the displacement direction is calculated by the following equation:

W = F · Δx

Where:

W = work.

F = applied force.

Δx = displacement.

In this case, we are only interested in the work done by the person and not in the net work done on the object (in which case we should consider the work done by gravity). And since the force is applied parallel to the displacement, the work done by the person will be:

W = 150 N · 7 m

W = 1050 J