Explore BrainMass

Explore BrainMass

    A Crate slides down a Ramp - Friction and Work

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A 50 kg crate slides down a 6 m ramp inclined at an angle of 35 degrees to the horizontal. A worker pushes on the crate parallel to the ramp so that it slides down with a constant velocity. If the coefficient of kinetic friction is .33

    a) Write equations of motion for the system

    b) Solve for the force exerted by the worker and compute the work done by the worker

    c) Compute the force of kinetic friction and find the work done by kinetic friction

    d) How much work is done by gravity?

    © BrainMass Inc. brainmass.com December 24, 2021, 7:17 pm ad1c9bdddf
    https://brainmass.com/physics/system-work/constant-velocity-friction-and-work-177140

    SOLUTION This solution is FREE courtesy of BrainMass!

    The solution file is attached.

    A 50 kg crate slides down a 6 m ramp inclined at an angle of 35 degrees to the horizontal. A worker pushes on the crate parallel to the ramp so that it slides down with a constant velocity. If the coefficient of kinetic friction is .33

    a) write equations of motion for the system

    b) solve for the force exerted by the worker and compute the work done by the worker

    c) compute the force of kinetic friction and find the work done by kinetic friction

    d) how much work is done by gravity.

    Ramp angle = θ = 35 deg, m = 50 kg, Length of the ramp L = 6 m, g = 9.8 m/s^2, μ k = 0.33

    (1) The crate has the following forces acting on it:
    Perpendicular to the incline: (a) Normal component of the self-weight W, which is W cos θ acting downwards and (b) The normal reaction force R acting upwards. These balance each other.
    ==> There is no motion perpendicular to the incline
    Parallel to the incline: (a) F (Pushing force acting down the incline) (b) the component of the self-weight W acting along the incline downwards, which is W sin theta and (c) Friction force Fk acting up the incline.
    This results in net motion down the plane at a constant velocity.
    ==> F = W sin θ - Fk

    (2) Self weight W = mg = 50 * 9.8 = 490 N
    Normal component of the self-weight W = W cos θ = 490 cos 35 = 401.38 N
    Normal reaction R = W cos θ = 401.38 N
    Component of the self-weight W acting parallel to the incline = W sin θ = 490 sin 35 = 281.05 N
    Force of kinetic friction Fk = μ R = 0.33 * 401.38 = 132.46 N
    F = W sin θ - Fk = 281.05 - 132.46 = 148.59 N
    Work done by the worker = Fd = 148.59 * 6 = 891.55 J.

    (3) Fk = 132.46 N
    Work done by friction = Fk * d = 132.46 * 6 = 794.76 J.

    (4) Work done by gravity = W sin θ * d = 281.05 * 6 = 1686.30 J.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:17 pm ad1c9bdddf>
    https://brainmass.com/physics/system-work/constant-velocity-friction-and-work-177140

    Attachments

    ADVERTISEMENT