# Diving car down steep, slippery hill (work, gravity, friction)

## Homework Statement

Driving an automobile down a slippery, steep hill, a driver brakes and skids at a constant speed for 10m. If the automobile mass is 1700kg and the angle of the slope of the hill is 25 degrees, how much work does gravity do on the car during the skid? How much work does friction do on the car?

ΣFx = 0

Wx = -mgsinΘ
Ff = ?

Ff - Wx = 0

ΣFy = 0

Wy = -mgcosΘ
N = mgcosΘ

N - Wy = 0

## The Attempt at a Solution

Ff - Wx = 0
Ff = Wx
Ff = -mgsinΘ
Ff = -(1700kg)(9.81m/s2)sin25°
Ff = -7.06x103N

It seems to me that if Ff is -7.06x103N then the gravitational pull down the hill would have to be an equal but opposite 7.06x103N, but I am unsure about this.

W = Fs cosΘ

Wg = 7.06x103N cos25° = 6.4x10 3J

Wf = -7.06x103N cos25° = 6.4x10 3J

I know that this is an incorrect answer, however, I am unsure of what to do.

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

Driving an automobile down a slippery, steep hill, a driver brakes and skids at a constant speed for 10m. If the automobile mass is 1700kg and the angle of the slope of the hill is 25 degrees, how much work does gravity do on the car during the skid? How much work does friction do on the car?

ΣFx = 0

Wx = -mgsinΘ
Ff = ?

Ff - Wx = 0

ΣFy = 0

Wy = -mgcosΘ
N = mgcosΘ

N - Wy = 0

## The Attempt at a Solution

Ff - Wx = 0
Ff = Wx
Ff = -mgsinΘ
Ff = -(1700kg)(9.81m/s2)sin25°
Ff = -7.06x103N

It seems to me that if Ff is -7.06x103N then the gravitational pull down the hill would have to be an equal but opposite 7.06x103N, but I am unsure about this.
You have correctly noted in your equations that since the car is moving at constant speed, then the sum of forces along the incline must be 0. So raise your hand if you're sure.
W = Fs cosΘ

Wg = 7.06x103N cos25° = 6.4x10 3J

Wf = -7.06x103N cos25° = 6.4x10 3J

I know that this is an incorrect answer, however, I am unsure of what to do.
What's the formula for work (your equations are incomplete)? Also, what is the value of theta to use in your equations for work?

You have correctly noted in your equations that since the car is moving at constant speed, then the sum of forces along the incline must be 0. So raise your hand if you're sure.What's the formula for work (your equations are incomplete)? Also, what is the value of theta to use in your equations for work?

I think I see my problem. I was thinking that since the car was going down a hill at at 25 degree angle that I would use the W = Fscos$$\theta$$ formula. However, after thinking about it, the force is parallel with the slope of the hill, so I should either use the W = Fx$$\Delta$$x formula OR set $$\theta$$ to 0 instead of 25.

PhanthomJay
Homework Helper
Gold Member
I think I see my problem. I was thinking that since the car was going down a hill at at 25 degree angle that I would use the W = Fscos$$\theta$$ formula.
the general equation for work done by a constant force is Fcostheta(d)
(However, after thinking about it, the force is parallel with the slope of the hill, so I should either use the W = Fx$$\Delta$$x formula OR set $$\theta$$ to 0 instead of 25.
yes, so how much work is done by gravity, and how much work is done by friction?

the general equation for work done by a constant force is Fcostheta(d)

OK, after relooking at my book what I thought was an Fs was actually an Fs, so W = Fscos$$\theta$$. Can I assume that s and d are the same thing?

yes, so how much work is done by gravity, and how much work is done by friction?

Wg = (7.06x103N)(10m) = 7.06x104J

Wf = (-7.06x103N)(10m) = -7.06x104J

PhanthomJay
Homework Helper
Gold Member
OK, after relooking at my book what I thought was an Fs was actually an Fs, so W = Fscos$$\theta$$. Can I assume that s and d are the same thing?

Wg = (7.06x103N)(10m) = 7.06x104J

Wf = (-7.06x103N)(10m) = -7.06x104J
Yes, yes, and yes to all. Now please study up on the definition of work...don't blindly copy a formula out of a book without trying to understand the theory.