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Homework Help: Work, power, and kinetic energy

  1. Sep 26, 2004 #1
    Hello, I recently arrived at a new school and I was put into a Physics class. I was in American History and English before this, so I was head first into a difficult subject with no basics down.

    I am currently working my hardest on these problems and need to know if I am doing it correctly.

    Question: What is the average power supplied by a 60.0 kg secretary running up a flight of stairs rising vertically 4.0 m in 4.2 s?

    Now, I know that I will be using power(P=W/T) and work(W=FD) equations, so I rewrote it as:

    P=W/T=Fd/t which also equals mgd/t

    so I then just substituted:

    (60.0kg)(9.81m/s)(4.0m)/4.2 s and since kg/m/s is J and J/s is W, the answer I got is 560.57 W.

    Question: A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an inclin at an angle of 20.0(degrees) with the horizontal. If there is no friction between incline and skates, what is the kinetic energy of the child at the bottom of the incline?(g=9.81 m/s(sup 2))

    So, I guess I am going to need kinetic friction equation KE=1/2mv(sup 2).

    Now I have the mass, but how do I go about finding the speed?(Hints please, no answers, I want to try and solve and understand :-))

    Question: A 40.0 N crate at rest slides down a rough 6.0 m long ramp inclined at 30.0(degrees) with the horizontal. The force of friction between the crate and ramp is 6.0 N. Find the velocity of the crate at the bottom of the slide.

    This one I am confused about. I don't even know how to start this one.

    Sorry for having such little understanding, but I am trying my hardest. Thanks for any help in adavnce!
  2. jcsd
  3. Sep 26, 2004 #2


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    Science Advisor
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    Dearly Missed

    Welcome to PF!
    Your first answer is right.
    Since there is no friction, mechanical energy is conserved.
    What does that mean?
  4. Sep 26, 2004 #3
    choose a reference frame :
    ramp = x-axis
    perpendicular to the ramp : y-axis

    A 40 N crate has a mass of 40/9.81 = 4.1 kg.

    now you need to calculate the component of gravity along the ramp. this component (draw a sketch here) : 40cos(-60°)

    Along the ramp the force of friction is : -6.0 (slows the object down thus - )

    Along the direction perpendicular to the ramp you have two forces (gravity and normal force which are of the same magnitude but the vectors have opposite sign thus no force in this direction.)

    the total force (along the ramp) : 40cos(-60°) - 6 = 4.1 * a. Solve this for acceleration a. Once you know a you can calculate the velocity v via :

    v = a*t.

    You need to know the time t of the movement along the ramp over 6 meters. this can be solved via x = x_0 + v_0 * t + at²/2 where x_0 is initial position (this 0 because the crate starts in the origin, we may chose this ourselves) and initial velocity v_0 is also 0. So 6 = at²/2 and solve for t...

    problem solved


    ps : try to figure out why I used an angle of -60° here (negative angle means start from the x-axis and go down the x-axis, clockwise...)
  5. Sep 26, 2004 #4
    as an addendum, in my example the ramp goes down when you start from the origin. You can also solve this when the ramp is inclined ofcourse. the angle becomes 60° then, this makes no difference though is cos(-60) = cos(60). this all depends on how you make your sketch...

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