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Let W1 represent the work required to accelerate an object on a frictionless horizontal surface from rest to a speed of v. if W2 represents the work required to accelerate the same object on the same surface from a speed of v to 2v, then which one of the following is the correct relationship between W1 and W2?

a) W2=W1 b) W2=2W1 c) W2=3W1 d) w2= 4W1 e) W2=5W1

Much appreciated.

Hi. I have this same question on an assignment due tomorrow as well! Since no one else has replied, I'll share what I've done (although I'm not sure if it is correct or not)
since equation of work is : w = fd (force x distance) and force = ma (mass x acceleration) we can substitute f with ma. So we write w = (ma)(d)
and acceleration is Vfinal - V initial (v is velocity)
SO
w1 = m(V-0)d and w2 = m(2v-v)d, this can be simplified to:
w1 = m(V)d and w2=m(V)d
so the answer would be w1=w2

The correct answer is (c). The total kinetic energy is four times higher at v2. The change is three times the value at v1.

Prof. Brian Zullkosky questions all over the web lol

Yup Phys 115 :/
Could you explain why the answer might be c)?

hdj
because when you double the speed you quadruple the kinetic energy
Ke = (1/2)m v^2
if w = 2 v
Ke = (1/2)m w^2 = (1/2)m (2 v)^2 = (1/2)m (4 v^2)

I understand how you get 4, but how would you get to the next step? How is it 3w1?

W1=ΔKE=KE2-KE1=mv²/2- 0= mv²/2
W2 = ΔKE=KE3-KE2=
=m(2v)²/2- mv²/2= 3mv²/2 =3W1

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