# Need Help with Work P

1. Nov 12, 2006

### cak

Need Help with "Work" P

Can anybody help me w/ this question??

A flight attendant pulls her 70.0 N flight bag a distance of 252 m along a level airport floor at a constant speed. The force she exerts is 44.0 N at an angle of 52.0° above the horizontal.

(a) Find the work she does on the flight bag.
(b) Find the work done by the force of friction on the flight bag.
(c) Find the coefficient of kinetic friction between the flight bag and the floor.

a) What I did was w=F*d*cos(angle) which would mean
w=44*252*(cos52)
w=-1807.241777 J
But it's on webassign and it told me that was incorrect...
Can someone tell me what I'm doing wrong??

b)I was kind of clueless on this one, but what I did was
Ffric=(mew)(m)(g)(cos.angle)=Work done for part a??

but since I dont have part A i couldnt finish this one or Part c....

2. Nov 12, 2006

3. Nov 12, 2006

### cak

haha i feel dumb :)
but that gave me the right answer

so for part c do i say...

6826.454=(mew)(7.14)(9.8)(cos52)?

4. Nov 12, 2006

Regarding (b) and (c) - the force of friction equals mu*N, but be careful about N. The weight is not the only force contributing to N; there is the applied pulling force, too! (i.e. its vertical component)

5. Nov 12, 2006

### cak

Does the angel contribute to the applied force??
If not, wouldnt N=44+70??

Also, Dont you need mu first in order to find the force of friction for part b??

6. Nov 12, 2006

### turdferguson

B is somewhat of a trick question because theres no acceleration. The forces in the x are equal and opposite

7. Nov 12, 2006

### cak

wait...cant you find acceleration because you know that w=mg
which means..70=m(9.8)
m=7.1429kg
and then f=ma
44=(7.1429)a
and you find that a=6.15996

but i dont understand what you need acceleration for