# Homework Help: Calculating work with kinematics and dynamics

1. Nov 22, 2011

### jjesiee

1. The problem statement, all variables and given/known data

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.
moving down at 4.0 m/s for 5.0 s?

2. Relevant 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 dont get?

2. Nov 22, 2011

### LabGuy330

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

Δd = (vf +vi)/2 * Δt

see if that doesnt give you 12.1644 kJ

3. Nov 22, 2011

### jjesiee

it wouldnt matter though cause vi= 0 (i think)

4. Nov 22, 2011

### sandy.bridge

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

5. Nov 22, 2011

### jjesiee

you sure? ive never heard of that before but thanks.

6. Nov 22, 2011

### 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.

7. Nov 23, 2011

### 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

8. Jul 16, 2013

### Enduro

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

fn +fg=ma

9. Jul 16, 2013

### PhanthomJay

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.