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Wow much thermal energy was generated by friction?

  • Thread starter starJ9
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  • #1
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Homework Statement


Suppose a car approaches a hill and has an initial speed of 116 km/h at the bottom of the hill. The driver takes her foot off of the gas pedal and allows the car to coast up the hill.

Randomized Variablesvi = 116 km/h
m = 780 kg
h = 20.5 m
a = 2.3 °

Part (a) If the car has the initial speed stated at a height of h = 0, how high (in m) can the car coast up a hill if work done by friction is negligible?

Part (b) If, in actuality, a 780-kg car with an initial speed of 116 km/h is observed to coast up a hill and stops at a height 20.5 m above its starting point, how much thermal energy was generated by friction in J?

Homework Equations


U2 + K2 = U1 + K1 + W_other
U2 is final potential energy.
K2 is final kinetic energy.

The Attempt at a Solution


Part (a) I got right.
1/2(mv^2) + mgh = 0
h = v^2/(2g)
=(32.2 m/s)^2 /(2*9.8m/s^2) = 52.9m

Part (b) I need help. (FK stands for friction force and s stands for displacement.)

I tried U2 + 0 = U1 + K1 + W_other.

W_other = -FK*s = U2 - U1 - K1
-FK*s = mg(Y2 - Y1) - 1/2(m*v^2)
-FK*s = 780kg*9.8m/s^2(20.5m*sin(2.3) - 52.9m) - 1/2(780kg*(32.2m/s)^2)
-FK*s = -652033j
FK = -(-652033j) / s
FK = 652033 j / (20.5m - 52.9m)
FK = -20124 N

The correct answer is 248200 j
 

Answers and Replies

  • #2
billy_joule
Science Advisor
1,200
330
What is the normal force?
What direction does the friction force act in?
Over what distance doe the friction force act?
 
  • #3
billy_joule
Science Advisor
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330
If you are familiar with conservation of energy you can get to the answer much faster..
 
  • #4
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What is the normal force?
What direction does the friction force act in?
Over what distance doe the friction force act?
The normal force would be mgcos(), 780kg*9.8m/s^2*cos(2.3) = 7638N.
The friction force acts 180 degrees from the car.
I'm assuming the distance is only 20.5m.

-FK*s = mgy - 1/2(mv^2)
-FK*s = 780kg*9.8m/s^2*20.5m - 1/2(780kg*32.2^2)
-FK*s = -248430 J
FK*s = 247666 J
 
  • #5
3,741
418
The 20.5 m is the height above ground. You don't need to multiply by any sine or cosine when you calculate the potential energy.
You don't need normal force. Just use energy balance, as it was suggested already.
Whatever energy is "lost" is converted into heat.
 
  • #6
billy_joule
Science Advisor
1,200
330
It seems I misread the question, the energy method is the only way to get the answer.

This threw me off:

W_other = -FK*s = U2 - U1 - K1
-FK*s = mg(Y2 - Y1) - 1/2(m*v^2)
-FK*s = 780kg*9.8m/s^2(20.5m*sin(2.3) - 52.9m) - 1/2(780kg*(32.2m/s)^2)
If μ was given you could've worked it out similar to your attempt (via W = Fd):

Wfriction = Fd
= μN * d
= μ (mg * cos (2.3) ) * (20.3m/sin(2.3))

eg heat generated equals friction force * the distance the cars odometer would read...the friction acts over a distance of greater than 500 metres. That's a lot farther than 20 metres!

-FK*s = mgy - 1/2(mv^2)
-FK*s = 780kg*9.8m/s^2*20.5m - 1/2(780kg*32.2^2)
-FK*s = -248430 J
FK*s = 247666 J
That looks correct.
 

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