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Posted by on Thursday, October 18, 2012 at 10:42pm.

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! - , Friday, October 19, 2012 at 12:26am

    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! - , Friday, October 19, 2012 at 1:51am

    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! - , Friday, October 19, 2012 at 2:57am

    Prof. Brian Zullkosky questions all over the web lol

  • Physics. please help! due tomorrow! - , Friday, October 19, 2012 at 3:04am

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

  • Physics. please help! due tomorrow! - , Friday, October 19, 2012 at 7:15am

    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! - , Friday, October 19, 2012 at 9:10am

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

  • Physics. please help! due tomorrow! - , Friday, October 19, 2012 at 11:34am

    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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