Work done and energy expended by a car traveling up a 30° slope

[pat666]

Hey i need some help with this question please:

A car of mass 2.135 * 10^3 travels at a constant speed up a slope of 30 degrees a distance of 25.3m. If the frictional resistance is 0.100 of the weight of the car, find:
(a) the work done
(b) the energy expended

The answer is 3.18 * 10^5 J

The formulas we have been using are w=fs and f=ma also si units please.

 

[rl.bhat]

What is the weight of the car?
What is the frictional resistance?
what is the component of the weight of the car along the slope?
What is the total force?

 

[pat666]

i can figure the weight out using f=ma but i keep getiing the wrong answer no matter what i do??

 

[pat666]

here' s what I've done so far.
weight = 2.135*10^4 N
frictional resistance = 2.094*10^3
and i try to put this into w=fs cos theata
and i get 4.589*10^5

please help

 

[rl.bhat]

Weight = 2.135*10^3*9.8 = 20923 N
The component of the weight of the car along the slope = 20923 N*sin(30 degree)
Total force = 20923 N*sin(30 degree) + 20923*0.100 N
Now find the work done.

 

[pat666]

hey thanks a lot for the help what i missed was that when you calculate the workdone it is the total force and not just the weight. thanks again