How Does Elevator Motion Affect Work Done by Forces?

[Enduro]

1. Homework Statement

the question is:
A 62 kg person in an elevator is moving up at a constant
speed of 4.0 m/s for 5.0 s. T / I C

(b) Calculate the work done by the normal force on the person.
(c) Calculate the work done by the force of gravity on the
person.
(d) How would your answers change if the elevator were
moving down at 4.0 m/s for 5.0 s?


2. Homework Equations
W= F x Δd

and i think

Δd= (vf + vi/2)Δt
fg=m.g
3. The Attempt at a Solution

for b) i did Δd= (vf + vi/2)Δt and got displacement which is 10 m and then i did fn=fg since fnet is 0 which means that they have the same force. and for that i got 607.6 N

p.s the answer in the back is 12 kJ which i don't get?

 

[voko]

At constant velocity v, what is displacement d over time t?

 

[PeterO]

1. Homework Statement

the question is:
A 62 kg person in an elevator is moving up at a constant
speed of 4.0 m/s for 5.0 s. T / I C

(b) Calculate the work done by the normal force on the person.
(c) Calculate the work done by the force of gravity on the
person.
(d) How would your answers change if the elevator were
moving down at 4.0 m/s for 5.0 s?


2. Homework Equations
W= F x Δd

and i think

Δd= (vf + vi/2)Δt
fg=m.g
3. The Attempt at a Solution

for b) i did Δd= (vf + vi/2)Δt and got displacement which is 10 m and then i did fn=fg since fnet is 0 which means that they have the same force. and for that i got 607.6 N

p.s the answer in the back is 12 kJ which i don't get?

The lift was moving at a constant speed, so calculating displacement is much simpler than the way you did it - and your method gives the wrong answer because you assumed it started out at 0 m/s.
This 5 second interval is presumably a middle interval of the whole journey.
The lift gains speed - then travels at constant speed - then slows down and stops. The 5 sec interval is entirely in the middle section.

 

[PhanthomJay]

1. Homework Statement

the question is:
A 62 kg person in an elevator is moving up at a constant
speed of 4.0 m/s for 5.0 s. T / I C

(b) Calculate the work done by the normal force on the person.
(c) Calculate the work done by the force of gravity on the
person.
(d) How would your answers change if the elevator were
moving down at 4.0 m/s for 5.0 s?


2. Homework Equations
W= F x Δd

and i think

Δd= (vf + vi/2)Δt
fg=m.g
3. The Attempt at a Solution

for b) i did Δd= (vf + vi/2)Δt and got displacement which is 10 m and then i did fn=fg since fnet is 0 which means that they have the same force. and for that i got 607.6 N

p.s the answer in the back is 12 kJ which i don't get?

I believe you meant Δd= [(vf + vi)/2)]Δt. If the speed is constant, vf and vi are the same. Try again.

 

[Nickie]

I would first commend the student for correctly using the equations for work and displacement. However, I would suggest that they also consider the direction of the forces and displacement in their calculations. For part (b), the normal force is acting in the opposite direction of the displacement, so the work done by the normal force would be negative. Similarly, for part (c), the force of gravity is acting in the same direction as the displacement, so the work done by gravity would be positive. This leads to the correct answer of 12 kJ for part (b).

For part (d), if the elevator were moving down, the direction of the forces and displacement would be reversed, resulting in a positive work done by the normal force and a negative work done by gravity. This would lead to the same final answer of 12 kJ. Overall, it is important to consider the direction of forces and displacement in order to accurately calculate work done in a system.