Wednesday
March 29, 2017

Post a New Question

Posted by on .

A 5.40 Kg package slides 1.47 meters down a long ramp that is inclined at 11.6 degress below the horizontal. The coefficient of kinetic friction between the package and the ramp is = 0.313. Calculate the work done on the package by friction. Calculate the work done on the package by gravity. Calculate the work done on the package by the normal force. Calculate the total work done on the package. If the package has a speed of 2.15 m/s at the top of the ramp, what is its speed after sliding the distance 1.47 meters down the ramp?

  • physics - ,

    component of weight normal to plane
    Fn = m g cos 11.6 = 51.9 N
    component of weight down the ramp
    Fd = m g sin 11.6 = 10.7 N

    Ff = Friction force = .313*Fn = 16.2 N

    work done by friction = -1.47 Ff = -23.9 Joules (force opposite to motion so negative)
    work done by gravity = m g * height change = m g * 1.47 sin 11.6
    = 15.7 Joules
    Normal force is perpendicular to motion so does no force
    Total work
    = 15.7 - 23.9 = - 8.2 Joules

    Ke at end = Ke at start -8.2 Joule
    (1/2) m v^2 = (1/2) m(2.15)^2 -8.2
    .5 m v^2 = 4.28 Joules
    v = 1.26 m/s

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question