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Homework Help: Kinetic Energy

  1. Nov 6, 2005 #1
    Can anyone help me with this?

    In a circus performance, a monkey on a sled is given an initial speed of 3.5 m/s up a 25° incline. The combined mass of the monkey and the sled is 17.0 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline does the sled move?

    I tried using the kinematic equations to find out the displacement. But I was left without the time. I know that KE= .5 x mass x velocity^2 but that isn't going to help me find out the displacement. I was also unsure of where to incorporate the incline angle. Thank-you
  2. jcsd
  3. Nov 6, 2005 #2
    You have the initial velocity.... 3.5m/s
    friction co-eff.... = 0.2
    Ff = -0.2FN
    You know that Fn is mgcos theta (Draw an FBD)
    ma = - 0.2 mgcos 25
    a = -0.2gcos25
    To find displacement, use the formula:
    v^2 = u^2 + 2as
    And solve for s...
    You have u, initial velocity, and v, final velocity. (= 0)
    And a...
  4. Nov 6, 2005 #3


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    Homework Helper

    You can solve this problem using either the kinematic eqns or by energy conservation method.

    Using the kinematic eqns, you could have used the eqn,
    v² = u² - 2as
    where v is the final velocity, u is the initial velocity, a is the deceleration, s is the distance travelled.
  5. Nov 6, 2005 #4
    to the a= -1.78 and the equation would be
    n 0^2=3.5^2+2(-1.78)x and it would come out to 3.45 but that answer is not correct when I enter into the answer checker
  6. Nov 6, 2005 #5


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    Homework Helper

    better not forget gravity!

    knowing (or wanting to know) distances is a task for Work & Energy.
    At the bottom, we have KE, while at the top we have PE_gravity.
    The process has friction force (m g cos(theta)) doing negative Work.
    Distance along the ramp is related to height uo the ramp by trig.
  7. Nov 6, 2005 #6

    Doc Al

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    Staff: Mentor

    The acceleration is not 1.78 m/s^2. (Pseudo Statistic's answer ignored the component of weight acting down the incline.)
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