What is the minuimum work needed to push a 950 kg car 180 m up along a 9.0 degree incline?
a) Ignore friction.
b) assume the effective cofficient of friction retarding the car is 0.25.

A. Work equals force applied times distance of travel. You know the distance, so all you need to determine is the force needed to push the car up the 9 degree incline. That would be the car's weight times the sine of 9 degrees.

B. Here the force needed is increased by friction, which is the coefficient of friction times the normal force of the car's weight against the incline.

Give this a shot and post your answers - we'll check them for you.

Part a ( ignoring friction)
F- mgsin(9)=0
so F= mg sin(9)
so W= F* d* cos(18o)= 950*9.80*sin(9)*810*cos(180)= -1179684J
Part b ( with friction)
Fn= mg cos(9)
Ffr= 0.25* mg cos(9)
so F+Ffr-mg sin(9)=0
F= mgsin(9)-(0.25*mg* cos(180)= -842.44J
so W= -842.44* 810= -682376.4J

Part a ( ignoring friction)
F- mgsin(9)=0
so F= mg sin(9)
so W= F* d* cos(18o)= 950*9.80*sin(9)*810*cos(180)= -1179684J


First - is the distance 810 (as you show here) or 180 (as you wrote in your first post)? Next, you shouldn't include the cos(180) term, because that gives a negative distance. That's why you got a negative number for W, but it should be positive. So use W = 950*9.8*sin(9)*810.


Part b ( with friction)
Fn= mg cos(9)
Ffr= 0.25* mg cos(9)
so F+Ffr-mg sin(9)=0


Should be F-Ffr-mgsin(9) = 0. Remember that the friction opposes the pushing, and so the force required to push the car is greater in part (b) than (a), not less.

A)
Firstly,the distance is 810 and not 180 as I wrote in the post.. it was a mistake!
Then,the angle in the formula W = F*d*cos(theata)
Is the angle between the force and the distance so it niether 180 nor 9 . It it is zero because F is the same direction of D..
b)
I totally agree.. and thank you for your note