Constant power, resistance, work, energy

  • Thread starter Demonfruzz
  • Start date
  • #1

Homework Statement


Mass car 1200 kg .

power car constant 20kw .

Resistance car motion 500N.

Car passes point X and Y with speeds 10 m/s ,25 m/s.

Car takes 30.5 s to travel from X to Y

1. Find acc of car at x
2.considering work/energy find distance XY


Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
T

Homework Statement


Mass car 1200 kg .

power car constant 20kw .

Resistance car motion 500N.

Car passes point X and Y with speeds 10 m/s ,25 m/s.

Car takes 30.5 s to travel from X to Y

1. Find acc of car at x
2.considering work/energy find distance XY


Homework Equations




The Attempt at a Solution

tried a few but looks nowhere near promising and I don't have an answer for the question ,past year examinations
 
  • #3
293
33
It is difficult to provide you any guidance if you do not show us your work. Please post your attempt at a solution, no matter how incorrect you may think it is.
 
  • #4
311
23
You might start by finding the net driving force at point A ( car driving force = power (watts) / velocity (m/s) )
The acceleration at that point = net force / mass

Hint: A velocity v acceleration graph covering the speed range will be a curve
 
  • #5
1. Distance used from question 2.
work = force x distance
Force =610KJ/470 m
=1297.9
acceleration = force/mass
=1297.9/1200
=1.0815

2. Power x time = work
Work= 20kw x 30.5
=610kj

Resistance = 500N x distancexy

Net =kinetic energy
610kj - (500N x distancexy )= (0.5 )
(1200)
(25^2)

Therefore distance xy = 470 m
 
  • #6
311
23
This looks like a calculus solution
 
  • #7
311
23
By short step excel based start and step program i get around 293 m distance
 
  • #8
You might start by finding the net driving force at point A ( car driving force = power (watts) / velocity (m/s) )
The acceleration at that point = net force / mass

Hint: A velocity v acceleration graph covering the speed range will be a curve

Hmm I see I will try to work it out , thanks for the help
 
  • #9
311
23
The program consides short velocity changes, adding the derived distance changes as it goes.
Consider the velocity change between 10 m/s and 11 m/s, calculate the (net) driving force at each velocity, find the average (net) force, calc the acceleration from : a = f net ave / mass, then use newons laws to calculate the distance travelled between those two velocities.
Add the steps as you go for a running distance.
Hint: i used steps of 0.01 metres / second, and 1500 ish lines of program
 
  • #10
Sorry am not really sure about calculus part
But maybe this is the solution

Driving force =1200*acc - 500N
P=fv
20kw=1200*acc - 500N
Acc=2.083

2.
Driving force =1200(2.083)- 500
=1999.6

Fs=0.5mv^2 - 0.5 mu^2
S={ [(0.5)(1200)(25^2)-[(0.5)(1200)(10^2)]} / driving force

Therefore distance =157.6m
 
  • #11
NascentOxygen
Staff Emeritus
Science Advisor
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2. Power x time = work
Work= 20kw x 30.5
=610kj

Resistance = 500N x distancexy

Net =kinetic energy
610kj - (500N x distancexy )= (0.5 ) (1200) (25^2)

Therefore distance xy = 470 m
You are almost right.

net work done = change in KE
and the car already has some KE at the beginning of that 30.5s time interval.
 
  • #12
293
33
Fs=0.5mv^2 - 0.5 mu^2

I don't think that this is correct, being as the force that propels the car forward is not constant.

Driving force =1200*acc - 500N
P=fv
20kw=1200*acc - 500N
Acc=2.083

A power output is not a force.
 
  • #13
Ao
You are almost right.

net work done = change in KE
and the car already has some KE at the beginning of that 30.5s time interval.


So the 2nd answer i gave should be the correct one ?

Thanks
 
  • #14
NascentOxygen
Staff Emeritus
Science Advisor
9,244
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So the 2nd answer i gave should be the correct one ?
No, because you introduced other mistakes there. :frown:

Fix the one I commented on.
 
  • #15
No, because you introduced other mistakes there. :frown:

Fix the one I commented on.

Yea I thought the 2nd answer was correct because it fixes the change in kinetic energy as you had pointed out
 
  • #16
haruspex
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But maybe this is the solution

Driving force =1200*acc - 500N
P=fv
20kw=1200*acc - 500N
Acc=2.083
you wrote, correctly, P=F v, but that's not what you did in the next step. What happened to v?
 
  • #17
you wrote, correctly, P=F v, but that's not what you did in the next step. What happened to v?

Owh sorry I made a typo I did include the v in the calculation
V = 10 at x
20k=(1200*accleration - 500 ) (10 ) Acc= 2.083
 
  • #18
haruspex
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Owh sorry I made a typo I did include the v in the calculation
V = 10 at x
20k=(1200*accleration - 500 ) (10 ) Acc= 2.083
Subtracting the 500N resistance made the acceleration go up. Does that seem reasonable?
 
  • #19
Thanks got it guys , conceptual problem
 

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