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Homework Help: Power/Work problem with some Newton's Second Law

  1. May 19, 2014 #1
    1. The problem statement, all variables and given/known data
    A 6.7-kg box is being lifted by means of a light rope that is threaded through a single, light, frictionless pulley that is attached to the ceiling.
    (a) If the box is being lifted at a constant speed of 2.0 m/s, what is the power delivered by the person pulling on the rope?

    Correct answer: 0.131 kW

    (b) If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.6 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope?

    Correct answer: 0.713 kW

    2. Relevant equations

    Power = Work/Time
    ∑F = ma

    3. The attempt at a solution

    For a, I've solved correctly;
    P=T*v = Tvcos(Θ)
    P=T*v (cos(0°) = 1, so I removed it)

    ∑F = ma
    a = 0
    ∑F = T - W = T- (m*g) = 0
    ∑F = T - (6.7kg*9.81 m/s^2) = 0
    T = 65.727 N
    P = 65.727 N * 2.0 m/s = 131.454 W = .131454 kW

    for b, I'm doing something wrong;

    I started with using the a kinematic equation to solve for a

    s = V(o)t + .5at^2

    1.6m/(.5*(0.42s)^2) = a = 18.14058957 m/s^2

    F = ma
    F = 6.7kg(18.14058957 m/s^2) = 121.5419501 N
    Work = |F||d|
    W = 121.5419501 N * 1.6 m = 194.4671202 J
    P = Work/time
    P = 194.4671202 J / .42 s = 463.0169528 W = .4630169528 kW

    Which is obviously wrong from the correct answer that is given....any help would be greatly appreciated. Thanks in advance.
    Last edited: May 19, 2014
  2. jcsd
  3. May 19, 2014 #2


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    Gold Member

    In part b when using newton 2, you have calculated the net force acting on the box. But the problem is looking power delivered by the man's force.
  4. May 19, 2014 #3
    Thanks a ton, that was all I needed, coming up with the right answer now.
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