# Calculate the work done by the mover

1. Apr 7, 2007

### raman911

1. The problem statement, all variables and given/known data
A mover pushes a 85kg crate along the floor at a constant speed through a displacement of 3,1m[E]. The coefficient of kinetic friction between the floor and crate is 0.22.
A) Calculate Fn and Fapp.
B] Calculate the work done by the mover.

3. The attempt at a solution
Fg=mg
=85kg*9.8N/kg
Fg=833N
So Fn=Fg
but how can i find Fapp

B] w=183N*3.1m
w=567.3

Last edited: Apr 7, 2007
2. Apr 7, 2007

### daniel_i_l

The box is moving at a constant speed, what does that say about the sum of the forces on the box?

3. Apr 7, 2007

### raman911

plz helpp.............

4. Apr 7, 2007

### arildno

What does the word constant mean??

5. Apr 7, 2007

not changing

6. Apr 7, 2007

### daniel_i_l

Yes, so what is the acceleration of the box? How can you use that to find the sum of all the forces?

7. Apr 7, 2007

### raman911

So a=0
Fnet=0
Fnet=Fapp+Ff
Fapp=0-180
Fapp=180
is that right?

Last edited: Apr 7, 2007
8. Apr 7, 2007

### raman911

$$Given$$

$$m=0.2kg$$
$$\Delta {h}_{1}=0.835m$$
$${v}_{1}=0m/s$$

$$Required$$

$${E}_{g} and {E}_{k}$$

$$Solution$$

$${E}_{g}={m}g\Delta h$$
$$={0.2kg}*9.8N/kg*0.835m$$
$${E}_{g}=1.63J$$

$${E}_{k}=(1/2)m{v}_{1}^2$$
$$=(1/2)0.2Kg(0m/s)^2$$
$${E}_{k}=0J$$

$${E}_{T}=1.56J$$

Last edited: Apr 8, 2007
9. Apr 7, 2007

### raman911

no response

10. Apr 7, 2007

### arildno

Where did you get the 180 from, and why did you cut out the sign in front of it?

11. Apr 7, 2007

### Hootenanny

Staff Emeritus
Your working seems ok, but I don't know where you got 180 from. Check your math...

Edit: Dammit, thats three times thats happened today :grumpy:

12. Apr 7, 2007

### raman911

soory i mean 183

13. Apr 7, 2007

### arildno

I couldn't care less if the number was 54 or square root of pi fifths.

I want you to tell us:

1. HOW did you arrive at that number??

2. WHY did you take away the sign?