Calculating work with kinematics and dynamics

[jjesiee]

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?

Homework Equations

W= F x Δd

and i think

Δd= (vf + vi/2)Δt
fg=m.g

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?

 

[LabGuy330]

what if i told you that your equation for Δd is incorrect.

Δd = (vf +vi)/2 * Δt

see if that doesn't give you 12.1644 kJ

 

[jjesiee]

what if i told you that your equation for Δd is incorrect.

Δd = (vf +vi)/2 * Δt

see if that doesn't give you 12.1644 kJ

it wouldn't matter though cause vi= 0 (i think)

 

[sandy.bridge]

The initial velocity is not vi=0, however, vf=vi since the velocity is constant.

 

[jjesiee]

The initial velocity is not vi=0, however, vf=vi since the velocity is constant.

you sure? I've never heard of that before but thanks.

 

[sandy.bridge]

Technically, if vi=0, then during that time frame there was indeed an acceleration and the velocity is then not constant. However, by equating vi=vf we are merely indication that the velocity remains constant and a=0.

 

[LabGuy330]

thank you sandy.bridge

vi does not equal 0

the statement says constant velocity of 4 m/s

thus vi = vf = 4 m/s

 

[Enduro]

the elevator is accelerating so fnet can not equal to 0.

fn +fg=ma

 

[PhanthomJay]

the elevator is accelerating so fnet can not equal to 0.

fn +fg=ma

No, it is given during that particular 5 second time interval that the elevator is not accelertaing, that is, it is given that it is moving at constant speed.