Work and friction for a 70 N flight bag pulled at 52° angle

[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 don't have part A i couldn't finish this one or Part c...

 

[radou]

Regarding (a) - set your calculator to 'degrees' instead of 'radians'.

 

[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)?

 

[radou]

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)

 

[cak]

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

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

 

[turdferguson]

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

 

[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 don't understand what you need acceleration for