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The question is: a horizontal force of magnitude 22.0 N is applied to a 2.86 kg book as the book slides a distance d = 0.73 m up a frictionless ramp at angle θ = 30°. The book begins with zero kinetic energy. What is its speed at the end of the displacement?

I did W=fdcos(theta) and found the amount of work to be 13.91J.

I then thought maybe you use the equation K=(1/2)mv^2
and rewrite as W=(1/2)mv^2
in which case v=3.12m/s

but this was wrong.

any guidance is appreciated! THANKS

  • physics -

    m=2.86 kg
    g=9.81 m/s/s
    θ=30 degrees
    F=22 N horizontal
    d=0.73 m (displacement)
    μ=0 (coefficient of friction)

    The weight of the mass can be resolved into two components,
    Normal to incline = mg cos(θ)
    along incline (downwards) = mg sin(θ)

    The horizontal force can be resolved into two components
    Normal to incline = F sin(θ)
    along incline (upwards) = F cos(θ)

    Net force pushing mass up incline
    = F cos(θ) - mg cos(θ)
    = 22 cos(30) - 2.86*9.81 sin(30)
    = 19.053 - 14.028
    = 5.024 N up incline

    Net acceleration up incline
    = 5.024 / 2.86
    = 1.757 m/s/s

    initial velocity = 0
    final velocity = v m/s
    distance = d = 0.73 m
    v2 - 02 = 2ad
    = 2*1.757*0.73
    = 2.565
    =1.60 m/s

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