Constant power, resistance, work, energy

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Homework Help Overview

The discussion revolves around a physics problem involving a car with a mass of 1200 kg, a constant power output of 20 kW, and a resistance of 500 N during motion. The car's speeds at two points, X and Y, are given as 10 m/s and 25 m/s, respectively, with a travel time of 30.5 seconds between these points. Participants are tasked with finding the acceleration of the car at point X and determining the distance between points X and Y using work and energy principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods to calculate acceleration and distance, including using net driving force, work-energy principles, and kinetic energy changes. Some express uncertainty about their attempts, while others suggest different approaches, such as using calculus or iterative methods.

Discussion Status

The discussion includes multiple interpretations of the problem, with participants providing hints and guidance without reaching a consensus. Some participants have shared their calculations and reasoning, while others have questioned the validity of certain approaches and assumptions made in the calculations.

Contextual Notes

Participants note the importance of considering initial kinetic energy and the impact of resistance on the calculations. There is also mention of potential errors in reasoning and the need for clarity in the application of formulas related to power and force.

Demonfruzz
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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

 
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Demonfruzz said:

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
 
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.
 
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
 
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
 
This looks like a calculus solution
 
By short step excel based start and step program i get around 293 m distance
 
dean barry said:
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
 
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 traveled 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
Demonfruzz said:
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
Demonfruzz said:
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.

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

A power output is not a force.
 
  • #13
Ao
NascentOxygen said:
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
Demonfruzz said:
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
NascentOxygen said:
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
Demonfruzz said:
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
haruspex said:
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*acceleration - 500 ) (10 ) Acc= 2.083
 
  • #18
Demonfruzz said:
Owh sorry I made a typo I did include the v in the calculation
V = 10 at x
20k=(1200*acceleration - 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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