Constant power, resistance, work, energy

1. Apr 25, 2015

Demonfruzz

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. The attempt at a solution

2. Apr 25, 2015

Demonfruzz

T
tried a few but looks nowhere near promising and I don't have an answer for the question ,past year examinations

3. Apr 25, 2015

AlephNumbers

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. Apr 25, 2015

dean barry

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. Apr 25, 2015

Demonfruzz

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. Apr 25, 2015

dean barry

This looks like a calculus solution

7. Apr 25, 2015

dean barry

By short step excel based start and step program i get around 293 m distance

8. Apr 25, 2015

Demonfruzz

Hmm I see I will try to work it out , thanks for the help

9. Apr 25, 2015

dean barry

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. Apr 25, 2015

Demonfruzz

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. Apr 25, 2015

Staff: Mentor

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. Apr 25, 2015

AlephNumbers

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

A power output is not a force.

13. Apr 25, 2015

Demonfruzz

Ao

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

Thanks

14. Apr 25, 2015

Staff: Mentor

No, because you introduced other mistakes there.

Fix the one I commented on.

15. Apr 25, 2015

Demonfruzz

Yea I thought the 2nd answer was correct because it fixes the change in kinetic energy as you had pointed out

16. Apr 25, 2015

haruspex

you wrote, correctly, P=F v, but that's not what you did in the next step. What happened to v?

17. Apr 25, 2015

Demonfruzz

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. Apr 25, 2015

haruspex

Subtracting the 500N resistance made the acceleration go up. Does that seem reasonable?

19. Apr 25, 2015

Demonfruzz

Thanks got it guys , conceptual problem

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