Download Example 4-10 Down the Slopes

Example 4-10 Down the Slopes
A 60.0-kg skier starts from rest at the top of a hill with a 30.0° slope. She reaches the bottom of the slope 4.00 s later.
If there is a constant 72.0-N friction force that resists her motion, how long is the hill?
Set Up
All three of the forces that act on the skier—the
s , the gravitational force w
s skier,
normal force n
and the friction force sf —are constant. So,
the net force on the skier is constant, as is her
acceleration. We can therefore use a constant-­
acceleration formula to find the length of the
hill. We’ll find the skier’s downhill acceleration
using Newton’s second law.
We choose the positive x axis to point
down the slope and the positive y axis to point
perpendicular to the slope as shown. (Compare
Example 4-6.) Then, the skier’s acceleration
points along the x axis only.
Solve
s has only a positive
The normal force n
s skier
y component. The gravitational force w
has a positive (downhill) x component and a
negative y component, while the friction force
has only a negative (uphill) x component. Use
these components and Newton’s second law
to solve for the skier’s x component of
acceleration ax.
1 2
a t (2-9)
2 x
Newton’s second law in component
form for the skier:
x = x0 + v0x t +
n
f
x: a Fext on skier, x = mskier ax
y: a Fext on skier, y = mskier ay
skier
y
O
Net force on skier = sum of normal
force, gravitational force, and
friction force:
x
O = 30.0°
s
s + w
s skier + sf
a Fext on skier = n
ny = n
Find the skier’s x component of
acceleration from the x component
of Newton’s second law:
a Fext on skier, x = wskier sin u + (2f )
= mskier ax
fx = –f
y
wskier, y = –wskier cosO
x
Weight of skier:
wskier
wskier = mskier g so
mskier ax = mskier g sin u 2 f
f
ax = g sin u mskier
= 19.80 m>s 2 2 sin 30.0 -
wskier
O
wskier, x = wskier sinO
72.0 N
60.0 kg
= 4.90 m>s 2 - 1.20 m>s 2
= 3.70 m>s 2
Now that we know the skier’s acceleration in
the x direction, we can use the kinematic equation to solve for her final position, which will
be equal to the length of the hill as long as she
starts at x0 = 0. Because the skier starts at rest,
her initial velocity is v0x = 0.
Reflect
Is our answer consistent? With an acceleration
of 3.70 m>s 2, our skier reaches a velocity of
14.8 m>s after 4.00 s. Because she started at
rest, her average x component of velocity for
the trip is half the final velocity, or 7.40 m>s.
If you travel at 7.40 m>s for 4.00 s, you cover
a distance of 29.6 m—which is just the answer
that we obtained.
Find the skier’s final position
after 4.00 s, assuming she starts
at x0 = 0.
1
x = x0 + v0x t + ax t 2
2
1
= 0 + 0 + 13.70 m>s 2 2 14.00 s2 2
2
= 29.6 m
y
x–
ent
m
x
ce
pla
dis
x0
x0
t=0
30.0°
x
t = 4.00 s
Skier’s velocity at t = 4.00 s:
vx = v0x + ax t
(2-5)
The skier’s initial velocity is v0x = 0, so
vx = 0 + 13.70 m>s 2 2 14.00 s2 = 14.8 m>s
Average velocity for the skier’s motion with constant
­acceleration:
vaverage, x =
0 + 14.8 m>s
v0x + vx
=
= 7.40 m>s
2
2
Distance traveled in 4.00 s:
x = x0 + vaverage, x t = 0 + 17.40 m>s2 14.00 s2 = 29.6 m
(2-7)