1. Hi, given this question in a physics lesson I am still having trouble competeing it. Can anyone help.

Calculate the change in kinetic energy of a car of mass 2000kg if it changes its velocity from 30kmh^-1 to 60kmh^-1.

Cheers,
Jack Heron
jeheron@gmail.com

2.

3. From my remote past unless things have changed

KE = Mass x velocity squared divided by 2g
g is engineers talk = [gravity 32.2ft /sec/sec]
hope that helps, you can do the math

4. WRONG!
its Ek=m(V<sub>1</sub>-V<sub>0</sub>)²/2
SI units

5. Um, no zelos

delta E = E<sub>end</sub> - E<sub>start</sub>

So in this case we have that

delta E = 1/2 m (v<sup>2</sup><sub>end</sub> - v<sup>2</sup><sub>start</sub>)

sub in m = 2000kg, v<sub>start</sub> = 8.333 m.s<sup>-1</sup> and v<sub>end</sub> = 16.666 m.s<sup>-1</sup>

So delta E = 208 333.333 J

6. Originally Posted by Zelos
WRONG!
its Ek=m(V<sub>1</sub>-V<sub>0</sub>)²/2
SI units
ita a basic formula, and the answer is in the question.
I am Die Fledermaus, eater of all things Zelos :wink:

7. First convert physical quantities in S.I syetem. You have to convert km/hr to m/s.

Change in KE=(v^2-u^2)m/2

v=Final velocity
u=Initial velocity
m=mass of the body

8. Originally Posted by Zelos
WRONG!
its Ek=m(V<sub>1</sub>-V<sub>0</sub>)²/2
SI units
change in KE=m(V<sub>1</sub>²-V<sub>0</sub>²)/2

9. Yes both Die Fliedermaus and Zelos goofed. OOPS. But river rat and anandsatya have it right.

KE = .5 m v^2

so you calculate KE for the two different velocities and subtract.

Just thought I should weigh in on this to help clear things up.

10. yeah youre river_rat. my bad

the real formula is thou
Ek=mc²(y-1)
y=1/sqr(1-v²/c²)

11. Originally Posted by Zelos

the real formula is thou
Ek=mc²(y-1)
y=1/sqr(1-v²/c²)
I never said a word.......

12. I never said a word.......
keep doing that

13. After some calculations I got the change in Ke to be 1800000kilojoules? Is this reasonable.

14. Originally Posted by jeheron
After some calculations I got the change in Ke to be 1800000kilojoules? Is this reasonable.
I don't think that's correct....

I get ~200 kJ.

Cheers,
wm

15. Ke = 1/2 * m * v^2

Kinetic energy pre acceleration (Ke1) = 1/2 * 2000 * (30*1000*60)
1/2 * m * velocity in m
= 18e9

Kinetic energy post acceleration (Ke2) = 1/2 * 2000 * (60*1000*60)
1/2 * m * velocity in m
= 3.6e9

Ke2 - Ke1 = 1800000000joules

= 1.8e6 kilojoules

which is my answer from the pervious post

16. Double check your conversion from km/h to m/s. That is where your error lies.

Cheers,
william

17. Kinetic energy pre acceleration (Ke1) = 1/2 * 2000 * (30*1000*60*60)
1/2 * m * velocity in m
= 108000000000

Kinetic energy post acceleration (Ke2) = 1/2 * 2000 * (60*1000*60*60)
1/2 * m * velocity in m
= 216000000000

Ke2 - Ke1 = 108000000000joules

=108000000kj

=108000mj

18. Check again....
Hint: km PER hour.

19. i am official a loser.

sorry.

stupid mistakes.

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