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Homework Help: Block, Ramp, Friction, and Spring work done

  1. Oct 15, 2008 #1
    http://img412.imageshack.us/img412/7050/blockspringrampfrictionfy0.gif [Broken]

    A man pulls a block of mass m = 20 kg up an incline at a slow constant velocity for a distance of d = 3.5 m. The incline makes an angle q = 33° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.2.

    a) What is the work Wm done by the man?

    b) What is the speed v of the block when it first reaches the horizontal surface?

    c) What is the spring constant k of the spring?

    d) How far up the incline d1 does the block rebound?

    Relevant equations:
    Wtotal = Change in Kinetic energy
    Ffriction = Coefficient of friction(Fnormal)
    Fspring = kx

    I know there are the force of friction, the man, and gravity on the box. I started by saying:
    Wman - Wgrav - Wfriction = Change in KE
    Wgrav = m*g(in x-direction)*(Change in height) -->(20kg)(9.81m/s^2*sin33)(3.5sin33m)
    Wfriction = (coefficient of friction)*m*g*distance -->(.2)(20kg)(9.81m/s^2)(3.5m)
    Change in KE = 0
    Wman = 823.919J, but this is not right...what am i doing wrong?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 15, 2008 #2
    What am i doing wrong with work done by the man? Isn't it just
    Wman-Wgrav-Wfriction = 0
    Wgrav = -m*g*h
    Wfric = -(Coefficient of friction)(m*g)(distance)
  4. Oct 15, 2008 #3


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    The work done by gravity is either -mgh or -mg(sin theta)*d. You used W_grav = -mg(sin theta)*d(sin theta) .

    Also, when calculating the friction force, the normal force and the weight are not the same.
  5. Oct 15, 2008 #4
    ok so your saying

    Wgrav = -(20kg)(9.81m/s^2)(sin33)(1.9m)

    And if your turn your coordinate axis so the normal force is going in the positive y direction wouldn't that make the normal force equal and opposite of the weight of the block?
  6. Oct 15, 2008 #5
    i figured it out

    Wgrav = m*g*(sin theta)*d
    Wfric = (Coefficient of friction)*(m*g*cos theta)*d

    after that it was quite easy to figure out the rest.

    Thanks PhanthomJay!
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