Physics. please help! due tomorrow!

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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.

  • Physics. please help! due tomorrow! -

    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

  • Physics. please help! due tomorrow! -

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

  • Physics. please help! due tomorrow! -

    Prof. Brian Zullkosky questions all over the web lol

  • Physics. please help! due tomorrow! -

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

  • Physics. please help! due tomorrow! -

    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)

  • Physics. please help! due tomorrow! -

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

  • Physics. please help! due tomorrow! -

    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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