Question

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.

Answer 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 <b>+</b> 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.

Answer 2

a=gsin(theta) not gcos(theta)