1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work and kinetic energy problem

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.51 m/s. The pulling force is 93.0 N parallel to the incline, which makes an angle of 20.1° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.08 m.

    a) How much work is done by the gravitational force on the crate?
    b) Determine the increase in internal energy of the crate-incline system due friction.
    c) How much work is done by the 93.0 N force on the crate?
    d) What is the change in kinetic energy of the crate?
    e) What is the speed of the crate after being pulled 5.08 m?

    2. Relevant equations
    Wgravity=-Δ(Ugravity)
    Ugravity=m*g*h
    W=F*Δr

    3. The attempt at a solution
    Please excuse lack of units.

    a) How much work is done by the gravitational force on the crate?
    Since the object is on an incline, the acceleration due to gravity must be adjusted:
    cos(20.1) * 9.81 = 9.21

    The height of displacement is also needed:
    sin(20.1) = x/5.08
    x=1.75

    Wgravity=-Δ(Ugravity)
    Wgravity=-(mghf-mghi)
    Wgravity=-(10.6*9.21*1.75 - 10.6*9.21*0)
    Wgravity= -171J

    I know that this answer is incorrect but do not know why.

    b) Determine the increase in internal energy of the crate-incline system due friction.
    Δ(KE) = Wfriction
    Wfriction= Fk * Δr
    Fkk * N
    Fk=0.400 * (9.21)(10.6)
    Fk=39.0
    Wfriction= 39 * 5.08
    W=198J=Δ(KE)

    c) How much work is done by the 93.0 N force on the crate?
    Wf=F*Δr
    Wf=93 * 5.08
    Wf=472J

    d) What is the change in kinetic energy of the crate?
    ΣW = Δ(KE) + Δ(Ug)
    Wfriction + Wf = Δ(KE) + Δ(Ug)
    -198J + 472 = Δ(KE) + 171J
    Δ(KE) = 103J

    e)What is the speed of the crate after being pulled 5.08 m?
    Δ(KE) = (1/2)*m*vf^2 - (1/2)*m*vi^2
    103 = (1/2) * 10.6 * vf^2
    Vf=2.10m/s
     
  2. jcsd
  3. Nov 24, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    The acceleration due to gravity is still 9.8 m/s². It is potential energy. It is a conservative field and gravity acts at all points in the downward direction against which it has moved to that height.

    Correct for the use of 9.21 elsewhere.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work and kinetic energy problem
Loading...