A 1250-kg car moves at 22.0 m/s. How much work is required from the engine to increase the car's speed to 36.0 m/s?

Question

rebeccaxhawaii · Answer

36/22

agent0smith · Answer

@rebeccaxhawaii not exactly...

Work done = change in kinetic energy

$$\large W = KE_f - KE_i$$

and KE = 0.5mv^2

idku · Answer

Yes, so,
0.5(1250)(36)^2-0.5(1250)(22)^2

idku · Answer

0.5(1250)(14)^2

idku · Answer

like this?

agent0smith · Answer

First step yes.

Second step no, very much no.

idku · Answer

Oh, yeah
36^2-22^2$\ne$14^2

agent0smith · Answer

You can not subtract squares like that.

36^2 - 22^2 is absolutely NOT equal to 14^2.

idku · Answer

I my bad ... that was my worst job factoring:)

rebeccaxhawaii · Answer

oops sorry its science related not really my thing but I do remember some

idku · Answer

tnx

idku · Answer

That is what I was getting wrong... that math :)

idku · Answer

Alright, I guess, meeting closed. Thanks for replying Smith!

agent0smith · Answer

You can factor out all the other stuff though

0.5(1250)(36)^2-0.5(1250)(22)^2 =

0.5(1250)(36^2-22^2)