Physics

A cyclist starts from rest and coasts down a 6.5∘{\rm ^\circ} hill. The mass of the cyclist plus bicycle is 95kg{\rm kg} . The cyclist has traveled 270m{\rm m} .What was the net work done by gravity on the cyclist? How fast is the cyclist going? Ignore air resistance.

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  1. Note:
    the degree symbol is written as
    & d e g ;
    (suppress all spaces)

    Difference in elevation, Δh
    = 270m × sin(6.5°)
    total mass (cyclist + bicycle), m
    = 95 kg
    total weight
    = mg

    Work done (on cyclist & bicycle)
    =mgΔh

    Component of force along slope, F
    = mg cos(θ)
    downward acceleration along slope, a
    = F/m
    = g cos(θ)

    Consider the acceleration over distance d of 270 m
    d=270 m
    vi=0 (initial velocity)
    vf (final velocity)
    a=gcos(θ)
    θ=6.5°
    g=9.8 m/s²
    m=95 kg

    We can apply the kinematics equation
    vf²=vi²+2ad
    to solve for vf.

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  2. a=gsin(theta) not gcos(theta)

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