The average rate of energy loss resulting from internal friction

[Franklie001]

Summary:: Having the following dates:

 

[BvU]

Interesting, but not so interesting that it merits posting twice !



Is there anything you want to ask ? To get help with homework, read this first

 

[Franklie001]

Hi,
sorry I'm new here.

Do you know how to find the average energy loss in the question posted by me at letter c?

I've found the work done= 44820J (996*45)

Kinetic energy of engine trasmission= 40500J

Hope you can help me.

 

[BvU]

question posted

Sure, but on my screen I can't read it. What is the problem statement, and what are the given data?

 

[anorlunda]

It does take time to learn the rules of a new place.

First, no duplicate threads (I'll delete the duplicate.). Second, homework goes in the homework forum (I moved it for you.) Third, to get homework help, you must first show us your work. Our helpers help. They don't do the work for you. That's what @BvU meant below.

So, don't get discouraged. Show us your work, and we'll try to provide some help.

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[Franklie001]

• A motor vehicle has a mass of 1·8 Tonnes, its wheels each have a moment of Inertia of 0·8kgm2 and their effective radius is 0·4m. The engine and transmission have an effective moment of Inertia of 0·9kgm2 [referred to the engine shaft]. The speed of rotation of the engine is 8 times that of the wheels.

The vehicle accelerates uniformly from 20m/s to 25m/s while covering a distance of 45m. During this time the average indicated power of the engine is 160kW and the effective road friction and air resistance is equivalent to a constant retarding force of 996N.
Determine:

• The total increase in kinetic energy.
• 
  
• The work done against road friction and air resistance.
• 
  
• The average rate of energy loss resulting from the internal friction in the engine transmission.
• 
  
• 
  
• 
  
• Engine & Transmission
  I = 0.9 Kgm2
  Engine Speed of Rotation = 8 x Wheel Speed
• Road Friction + Air Resistance = 996 N
• Mass of Vehicle
  1.8 Tonne
• Wheels
  I = 0.8 Kgm2
  r = 0.4 m
• Other Data
  
  Initial Velocity = 20 m/s
  Final Velocity = 25 m/s
  Distance = 45 m

 

[Franklie001]

I've already found the first question with 245.25*10^3J and the second at 44820 N*m by the way

 

[BvU]

245.25*10^3J

That's strange: I found something different. Can you post your work ?

[edit]never mind, I still have something left to do . Hold on...
[edit2]No: tried left and right (i.e. include change in kinetic energy of rotation as well )
but can't find what you found. In sucha case we always ask: post what you did ...

And a small comment: use a space between a number and the unit
And another small comment: you were given data in 1 or 2 digits, so answers (*) should be in 2 or max 3 digits

(*) only the answer you hand in. Intermediary results should not be rounded off.

 

[Franklie001]

Ok let's start from question 1:

For question 1, I calculated all the difference of kinetical energy of vehicle, engine- trasmission and 4 wheels.

KE Vehicle
u=20 m/s , v=25 m/s

KE= 1/2 * v^2 * m - 1/2 * u^2 * m
KE= (1/2 * 25 ^2 * 1.8 x 10^3) - (1/2 * 20^2 * 1.8 x 10^3) = 202.5 * 10^3 J

Four Wheels KE
Inertia=0.8 kg m^2 (one wheel)
r=0.4m

Inertia for 4 wheels= Inertia x 4 = 0.8 * 4=3.2 kg m^2

Rotational velocities convertion:
W2= v2/r = 25/0.4 = 62.5 rad/s
W1= v1/r = 20/0.4= 50 rad /s

KE=(1/2 * (I * 4) * W2 ^2) - (1/2 * (I*4) * W1 ^2)
KE= (1/2 * 3.2 * 62.5 ^2) - (1/2 * 3.2 * 50 ^2)
KE= 6250 - 4000 = 2250 J

Engine Trasmission
I=0.9 kgm^2

Speed of rotation= 8 * Wheel speed
W2 trasm.= 8 * W2 = 8 * 62.5= 500 rad/s
W1 trasm.= 8 * W1= 8 * 50= 400 rad/s

KE= (1/2 * I * W2 ^2) - (1/2 * I * W1^2)
KE= (1/2 * 0.9 * 500^2) - ( 1/2 * 0.9 * 400^2)
KE= 112500-72000= 40500 J

Total kinetic energy= (202.5 * 10^3) + 2250+ 40500 = 245 x 10^3 J

 

[Franklie001]

b) Work done= Force x distance

Work= 996 * 45= 44820 N*m

c) For question c I tried different ways calculating powers and subtract them but I still can't get the right result.

Any advice?

Thanks for your help

 

[BvU]

You're doing better than I am ! Finally reproduced your a) result (had a typo in excel...)..

still can't get the right result.

Apparently you already know what is the right result.

What did you do te almost get it ? Please post

The average rate of energy loss resulting from the internal friction in the engine transmission.

I suppose the wheels can be ignored, or else they mean 'including the wheels'...

So you need an energy input, the kinetic energy increase from a) and a delta time ...

 

[Franklie001]

Hi,

Sorry what you mean by I need an energy input. kinetic energy from a and delta time?

Which formula should I use?

 

[Franklie001]

Honestly I started calculating the input power finding the angular acceleration and the torque after but after I time it with the inertia given I don't get the result with is 14.965 KW

 

[BvU]

The power input is given

During this time the average indicated power of the engine is 160kW

It is used to add kinetic energy (a), to overcome road friction and air resistance (b) and the remainder is (c).

All you need to find out to convert your 245250 J and and 44820 J to kW. How does one do that ?

 

[Franklie001]

How do I convert joules to kW?

 

[Franklie001]

I am really sorry to bother you.

I appreciate your help by the way

 

[BvU]

How do I convert joules to kW?

$$ 1 \ {\sf W} = 1 \ {\sf J/s}$$

160 kW means 160000 Joules per second

To convert your 245250 J change in kinetic energy to kW, and idem the 44820 J, you need to know how long the acceleration lasted.
For that, you have two givens: the change in speed and the distance covered. Need I say more ?

And you are not bothering anyone ! We are here to help.

 

[Franklie001]

Do I need to sum the two converted energy? And what after?

 

[Franklie001]

Will you be able to provide me the steps to use to find the result, with the right formula?

Thanks

 

[BvU]

Yes I can, but that doesn't help you. I am trying to guide you to help you find the answer yourself.
So please answer:

Is it clear to you hat power and rate of energy loss have the same dimension: energy/time, so that, if you have the time you can determine the (average) power ?

Can you find out how long the acceleration lasted ?

 

[Franklie001]

Thank you

I found the right solution