Does an Elevator Affect Your Weight?

[tennisacerg]

A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increases to 1050 N, then returns to normal. When the elevator slows to a stop, his weight drops to 588 N. Determine the acceleration at the beginning and end of the trip.

How do you solve this?

Please help!

 

[PhanthomJay]

A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increases to 1050 N, then returns to normal. When the elevator slows to a stop, his weight drops to 588 N. Determine the acceleration at the beginning and end of the trip.

How do you solve this?

Please help!

As the elevator rises or slows
$$F_{net} = ma$$ where $$F_{net}$$ consists of his weight acting down, and the normal force acting up. What is the person's weight? What is the normal force acting up on his feet? Solve for a in each case. Watch your plus and minus signs.

 

[kyle8921]

I would approach this question by using the fundamental principles of physics, specifically Newton's laws of motion. The first law states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. Finally, the third law states that for every action, there is an equal and opposite reaction.

In this scenario, the student's weight is being affected by the forces acting on him in the elevator. When the elevator is at rest, the student's weight is his normal weight of 840 N. This is because there is no net force acting on the student, so his weight is equal to the force of gravity pulling him towards the center of the Earth.

As the elevator rises, the student's weight increases to 1050 N. This is because the elevator is accelerating upwards, and the student is experiencing an upward force from the elevator floor. This upward force is greater than the force of gravity pulling him down, resulting in an increase in weight.

When the elevator slows to a stop, the student's weight drops to 588 N. This is because the elevator is decelerating, and the student is experiencing a downward force from the elevator floor. This downward force is less than the force of gravity pulling him down, resulting in a decrease in weight.

To determine the acceleration at the beginning and end of the trip, we can use the second law of motion. We know that the student's mass remains constant, so the acceleration is directly proportional to the net force acting on him. At the beginning of the trip, the net force is the difference between the force of gravity and the upward force from the elevator floor, which is 1050 N - 840 N = 210 N. So, the acceleration at the beginning of the trip is 210 N / student's mass.

At the end of the trip, the net force is the difference between the force of gravity and the downward force from the elevator floor, which is 588 N - 840 N = -252 N. Since the student is experiencing a downward force, the acceleration at the end of the trip is -252 N / student's mass.

In summary, the acceleration at the beginning of the trip is 210 N / student's mass, and the acceleration at the