Question on work energy and power kinetic energy of an ....

Click For Summary

Homework Help Overview

The discussion revolves around a problem in the context of work, energy, and power, specifically focusing on the kinetic energy of a vehicle as its speed increases. The original poster presents a scenario where the speed of a vehicle increases by 2 m/s, leading to a doubling of its kinetic energy, and seeks to find the original speed of the vehicle.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the kinetic energy formula and the implications of the speed increase on kinetic energy. There are attempts to derive the original speed using algebraic manipulation and the quadratic formula. Some participants question the correctness of the calculations and the application of the quadratic formula.

Discussion Status

The discussion is ongoing, with participants revisiting their calculations and correcting mistakes. There is a recognition of errors in the application of the quadratic formula, and some participants express gratitude for the assistance provided in identifying these mistakes. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the original speed derived does not match the provided options, leading to further scrutiny of the calculations. There is an emphasis on ensuring the correct application of mathematical principles without reaching a definitive conclusion.

Mr.maniac
Messages
38
Reaction score
0

Homework Statement


If the speed of a vehicle increases by 2m/s then it's kinetic energy is doubled. Find out the original speed of the vehicle

Homework Equations


1/2(mv^2)

The Attempt at a Solution


First find out both KE(kinetic energy) taking u(initial vel.)
And v(final vel.) And mass as m
K(initial KE)= 1/2(mu^2)
K`(final KE)=1/2(mv^2)
=1/2[m(u+2)^2]. {Given "speed of vehicle increases by 2m/s}

Also
2K=K`
Then equating
2[1/2(mu^2)]=1/2[m(u+2)^2].
=>mu^2=1/2(mu^2+4m+4mu)
=>u^2=1/2(u^2+4+4u) {dividing m on both sides}
=>2u^2=u^2+4+4u
=>0= -u^2+4+4u.
{Further using quadratic formula}
We get
u=(√2+1)÷2
Sadly that ain't there in the options so. :~( {booo}
 
Physics news on Phys.org
Mr.maniac said:

Homework Statement


If the speed of a vehicle increases by 2m/s then it's kinetic energy is doubled. Find out the original speed of the vehicle

Homework Equations


1/2(mv^2)

The Attempt at a Solution


First find out both KE(kinetic energy) taking u(initial vel.)
And v(final vel.) And mass as m
K(initial KE)= 1/2(mu^2)
K`(final KE)=1/2(mv^2)
=1/2[m(u+2)^2]. {Given "speed of vehicle increases by 2m/s}

Also
2K=K`
Then equating
2[1/2(mu^2)]=1/2[m(u+2)^2].
=>mu^2=1/2(mu^2+4m+4mu)
=>u^2=1/2(u^2+4+4u) {dividing m on both sides}
=>2u^2=u^2+4+4u
=>0= -u^2+4+4u.
{Further using quadratic formula}
We get
u=(√2+1)÷2
Sadly that ain't there in the options so. :~( {booo}
Check your work in using the quadratic formula.
 
Will do
 
OK redid it made a slight mistake but it's still no good

u=(-1+/-√2)÷2
 
The +/- means plus or minus or this ± sorry didn't notice the sign before
 
Mr.maniac said:
OK redid it made a slight mistake but it's still no good

u=(-1+/-√2)÷2
That's also wrong.

Show your steps.

(The initial try looked better.)
 
Mr.maniac said:
=>0= -u^2+4+4u.
{Further using quadratic formula}
We get
u=(√2+1)÷2
Your quadratic looks good, but redo your application of the quadratic formula.

(Edit: SammyS beat me to it!)
 
OK got the mistake the ans now is 2(√2+1)
Which is correct I was taking the denominator wrong thank you very much for pointing that out
 
Thank you all for helping
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
6
Views
1K
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 58 ·
2
Replies
58
Views
3K
Replies
7
Views
2K
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K