# Calculate Energy Lost to Friction After 100m for 600 kg Car on 15 Deg Hill

• physiman
In summary, a car with mass 600 kg accelerates uniformly from rest down a steady hill inclined at 15 degrees to the horizontal. The average frictional force acting on the car as it free-wheeled down the hill was 152340.9 J.

#### physiman

First, a big hello to the Physics Forums community :)

A car with mass 600 kg accelerates uniformly from rest down a steady hill inclined at 15 degrees to the horizontal[...]
- Work out the energy lost to friction after the car has traveled 100m
- The average frictional force acting on the car as it free-wheeled down the hill

For the first 3 parts I had to work out the speed of the car after 100m in 20s, the KE after the car has traveled 100m and the GPE the car has lost after tavelling 100m down that slope. I got 10 ms^-1; 30 000 J and 152340.9 J respectively.

So I tried using the Work done formula to out the energy lost due to friction:

W=Fd

For W I used the GPE the car has lost in PE, so 152340.9 J and got the value of F to be 1523.4 N. Is that the right method, I'm unsure because I think that I have to include the Kinetic Energy as well but the difference between KE and GPE looks too big for me.

Or do I have to use the F=ma formula? By that I mean work out a from the values I have, 0.5 m s^-2, and then I get 300N as the value for F but that doesn't look right to me as it has nothing to do with the energy loss.
well, I guess I can got to the next question only after I solve this one.

Was the time given in the problem?

Yes, sorry forgot to mention that.
The time given was 20s.

Is the engine off?

That wasn't mentioned, it says it 'free-wheeled' down the hill.

Would that mean the engine was off?

I think so.

BTW try to check your working for the GPE.
Are you taking 9.8 for g?

9.81 to be precise ;)

What is your working for the GPE?

Sorry - Your GPE is ok .
i was using 9.8 instead of 9.81/

I used mgh.

But I worked out the height from the ground, because GPE depends on it.

So I just used 100sin15 and got about 25.9 m for it.

I got F = 1223.41N

Did you use W=Fd?

Because my answer (1523.4 N) doesn't look far away from your one.

Yes I used W = Fd.

Note that when you got 300 for the force, that was the RESULTANT or NET force.
You CAN use Fnet = ma and you still get 1523N for the frictional force.

How can I get to the Friction force from working out the resultant force?

Oh yh, and about the energy lost to friction bit:

Do I just work out the Net energy from before motion and after the car has traveled 100m?

i.e.:

Initial KE: 0 J
Initial GPE: mgh = 152340.8899...J

Total Energy before moving: 152340.9 J

After traveling 100m:

KE = 1/2mv^2 = 30000 J
GPE = mgh = 0 J (assuming that it has reached the horizontal)

Therefore making the energy lost to friction:

152340.8899... - 30000 = 122340.8899... J

E lost to friction is ok
re friction from net force:
How many forces act along the track?

Not mentioned on the Question but the only forces I can think of is Weight of the car (perpendicular to the track, therefore 600(9.81)cos15 ), the frictional force and there should be a Normal reaction too.

I mean forces ALONG THE TRACK. 600(9.81)cos15 ) is the component of the weight perpendicualr to the track and the normal reaction is also perpendicular to track.
Along the track there is the friction and the ...

oh alright, sorry.

tbh apart from the friction I can't think of any force, things like air resistance should be negligible

Correct. Air resistance is being ignored but there is a component of the weight downward along the track opposing the friction upwards along the track. Right?

Ok, that must be the Horizontal force then, so 600(9.81)sin15 instead of using cosine?

exactly!
Now you can find the friction form the NET force along the track.

Oh cool, I did get ~1523.41 N and from the 1223.41 worked out for friction it would support F=ma.

Thanks a lot, you were a real help. If there was a reputation system I would show my appreciation :)

You are always welcome to this forum.

## 1. How do you calculate energy lost to friction after 100m?

The formula for calculating energy lost to friction is: Energy lost = force of friction x distance. In this case, the force of friction can be calculated using the formula: force of friction = coefficient of friction x weight of the car. Once you have the force of friction, you can multiply it by the distance traveled (100m) to get the total energy lost to friction.

## 2. What is the coefficient of friction for a 600 kg car on a 15 degree hill?

The coefficient of friction varies depending on the surface and conditions, but for a standard road surface, the coefficient of friction for a car is typically between 0.03 and 0.06. For the purposes of this calculation, we will use the average of 0.045.

## 3. How does the weight of the car affect the energy lost to friction?

The weight of the car directly affects the force of friction, which in turn affects the energy lost to friction. The heavier the car, the greater the force of friction, and therefore the more energy will be lost to friction over a given distance.

## 4. Why is the angle of the hill important in calculating energy lost to friction?

The angle of the hill affects the force of friction by changing the normal force on the car. The steeper the hill, the greater the normal force, and therefore the greater the force of friction. This means that more energy will be lost to friction on a steeper hill compared to a flatter one.

## 5. Is this calculation accurate for all types of roads and conditions?

No, this calculation is based on the average coefficient of friction for a standard road surface and assumes a constant coefficient of friction over the entire distance. In reality, the coefficient of friction can vary and different road conditions, such as wet or icy roads, can greatly affect the energy lost to friction. This calculation should be used as an estimate and may not be entirely accurate in all cases.

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