How much work is done by friction during the car's slide down the hill?

[fatty1]

Homework Statement

A 320 kg car slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the car and the hill surface is 0.23.

1) If the gravitational potential energy PE of the car-Earth system is set to zero at the bottom of the hill, what is the value of PE just before the slide?

2) How much mechanical energy is wasted by frictional forces during the slide?

Homework Equations

PE = mgh

Frictional force = (kinetic friction) * (mg cos theta)

Work = Final Energy - Initial Energy

The Attempt at a Solution

I tried to solve the first question with the potential energy equation but I am having trouble solving the second equation.

 

[Doc Al]

I tried to solve the first question with the potential energy equation but I am having trouble solving the second equation.

I assume you can treat the hillside as if it were a straight incline.

What's the force of friction? You have the formula. What's the angle of the incline?

Use that friction force to calculate the work done by friction.

 

[fatty1]

I assume you can treat the hillside as if it were a straight incline.

What's the force of friction? You have the formula. What's the angle of the incline?

Use that friction force to calculate the work done by friction.

I did it this way.

1) PE = mgh = 320 x 9.8 x 300 = 940800 J

2) By work energy theorem:
=>W = Delta Energy
=>Delta Energy = kinetic friction * distance
=>Delta Energy = 0.23 * 320 * 9.8 * [300/500] * 500
=>Delta Energy = 216384 J


I did it this way, but for question 2, should the answer be negative since it loses energy?

 

[Doc Al]

I did it this way.

1) PE = mgh = 320 x 9.8 x 300 = 940800 J

Perfectly correct.

2) By work energy theorem:
=>W = Delta Energy
=>Delta Energy = kinetic friction * distance
=>Delta Energy = 0.23 * 320 * 9.8 * [300/500] * 500
=>Delta Energy = 216384 J

You're using the definition of work-- work=force*displacement--not the work-energy theorem. You've made an error: 300/500 is the sine of the angle, not the cosine.

I did it this way, but for question 2, should the answer be negative since it loses energy?

The work done by friction is negative, since the force and the displacement are in opposite directions. But the amount of energy 'wasted' would be positive.

 

[Sakriya]

potential energy just before the slide(at top of hill)= mgh= 320*10*300

let me give you some help with the second question. If the energy were not to be lost then the whole of the potential energy would have converted into kinetic energy and velocity at the bottom would have been v=sqrt(2gh) .but in this case the velocity will be less than this calculated velocity and this is where the energy is lost.

So at first calculate the acceleration down the hill of the car.

ma= mgsinA -(Coeff of friction, U)*mgcosA , I am using A in place of theta
a=4.16 m/s^2 if you put g=10m/s^2

Now velocity at the bottom is given by
v^2=2*a*S(hypotenuse distance of the hill)
v=64.5 m/s

therefore K.E at bottom =665600 J
Loss is energy = P.E at top - K.E at bottom

 

[fatty1]

2)
=>W = Delta Energy
=>Delta Energy = kinetic friction * distance
=>Delta Energy = 0.23 * 320 * 9.8 * [500/300] * 500
=>Delta Energy = 601067 J

I reversed that and got this value. If it is not correct, then can you explain much clearer about the work energy theorem?

 

[Doc Al]

2)
=>W = Delta Energy
=>Delta Energy = kinetic friction * distance
=>Delta Energy = 0.23 * 320 * 9.8 * [500/300] * 500
=>Delta Energy = 601067 J

I reversed that and got this value. If it is not correct, then can you explain much clearer about the work energy theorem?

Again, this is not the work-energy theorem. You are just using the definition of work.

You still made an error in calculating the friction force, which should be μmg*cosθ. Before you had sinθ, instead of cosθ; now you have 1/sinθ, which is still not correct. Hint: In the right triangle formed by the hill, the hypotenuse is 500, the side opposite the angle is 300, so what's the third side?

FYI, the work-energy theorem states that the net work done on a body will equal its change in energy. We don't need to use that here. You could use it, but doing it the way you are attempting to do it is better and quicker. You just need to get the correct value for cosθ.