Variable formula question Why does W2 = 24W1?

[Lo.Lee.Ta.]

Variable formula question... Why does W2 = 24W1?

1. "To accelerate a certain car from rest to the speed v requires the work W1. The
work needed to accelerate the car from v to 5v is W2. Which of the following is correct?"

2. The only forumulas I can think of that deal with work are:
W = Fd

F = ma

a = v/t

...So if you tried to replace the F with velocity terms, you could get:

W1 = m(v/t)d

...Now they say that W2 = [m(5v/t)d] - [m(v/t)d]
They say W2 is the the work needed to accelerate v *TO* 5v, so is that right...?

So W2 = m(4v/t)d, yes? :/

So, if I switch the v/t back to force, then:

W2 = m(4a)(d)

W2 = (F/4)(d)

So W1 = (F)(d) and W2 = (F/4)(d)

Or, 4W2 = Fd = W1

So, it seems like 4W2 = W1 BUT THIS IS COMPLETELY NOT RIGHT!

=_= I don't get this problem at all! Help me, somebody!
The answer's supposed to be W2 = 24W1 !
UGH!

Thank you!

 

[WillemBouwer]

W2 = m(4a)(d)

W2 = (F/4)(d)

If you say F = ma how do you say m(4a) is F/4? That is not correct...

Also importantly, how is a = (v/t)... If you look at your uniform linear motion equations.
v = u +at so that says a = v/t for the first part but what is the acceleration for the second part? Is it uniform? Try figuring out your problem here?

 

[WillemBouwer]

Lo.Lee.Ta: Okay let's assume the acceleration is uniform.
You stated that W1 = Fd = mad what is the equation for W_2? remember the distance is not the same...
What equation can you use to get the distance travelled? Linear equation?
Also what is the time it takes for the second part if we say that the acceleration is constant?
If you substitute these into the right equations you will see that W2 = 24W1... but reply to my questions and I will help you further...

 

[Lo.Lee.Ta.]

...
W1 = Fd = ma x d

W2 = Fd = ma x d

I'm having trouble relating the 5x velocity to the equations...
Can we say that if velocity is 5x greater, then the distance for W2 is 5x greater?

...This is the only thing I can think of. :/

 

[ap123]

Can't you just use the work-kinetic energy theorem?

 

[superdave]

The Work-Kinetic Energy Theorem states that Work = Change in KE. Try going from there.

 

[Lo.Lee.Ta.]

Okay, so the Work-Energy Theorem is:

Wtotal = ΔKE = 1/2mvf^2 - 1/2mvi^2

= 1/2(m)(5v)^2 - 1/2(m)(v)^2
= (1/2(m)v^2 x 25) - (1/2(m)v^2 x 1)
Work(total) = 1/2(m)v^2 x 24

Okay, I think I am seeing where the number 24 comes from...
So is this meaning that the [1/2(m)v^2] stands for W1?

So, yeah, that would be 24W1 = W2.
Is this the right reasoning?

THANK YOU, GUYS! :D
You all are awesome!

 

[WillemBouwer]

Yes that is correct, I just wanted to show you that the equation you have also works:
with W1 = mad_1 and W2 = mad_2
with d_1 = ut +0.5at^2 = 0.5vt
and with the acceleation being constant the time for the second period:
v/t_1 = (5v-v)/t_2
so t_2 = 4t_1
hence d_2 = ut + 0.5at^2
d_2 = (v)(4t) + 0.5(v/t)(4t)^2
d_2 = 12vt

Substituting gives you W_1 = m(v/t)(0.5vt) = 0.5mv^2 (here you can see where the KE equation comes from)
W_2 = m(v/t)(12vt) = 12mv^2 so this will be 24*W_1... So all in all the KE equations is much easier but your approach was not wrong it gets to the same answer...