Kinetic Friction

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Transcript Kinetic Friction

Kinetic Friction
Sliding

Sliding objects also have a frictional force exerted on
them.

This frictional force is kinetic friction.

An approximate formula:
F fr  m k FN
mk is the coefficient of kinetic friction
Continuing Motion


FN  mg cosq
• Fgx = mg sinq.
• Block increases or
decreases speed depending
on which is greater.
F fr  m k mg cos q
mg sin q
q
mg cos q
Ffr = mkFN = mkmg cosq
Compare to the downslope
component of gravity

If equal there is constant
velocity.
Fg  mg
mg sin q  m k mg cosq
m k  tan q
Static vs. Kinetic

Static and kinetic friction are
similar.

Static and kinetic friction are
different.
• Force in opposite direction
to motion
• Static friction is an inequality
up to a maximum
• Proportional to normal force
• Coefficient of friction is
typically greater for static
friction at the same surface
• Coefficient of friction
depends on materials
• Wood on wood (0.4 vs. 0.2)
Drag

Kinetic friction is a constant force.
• If there is a net force an object would accelerate forever

Air resistance causes a friction called drag.
• At low velocity drag is proportional to velocity, Fd = -bv
• At higher velocity drag goes as velocity squared, Fd = -cv2

Because it depends on velocity the force increases
as an object speeds up.
Terminal Velocity

An object may fall through
the air at constant velocity.
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By the law of inertia the net
force is zero.

The force of drag must
balance the force of gravity.
Fd = cv2
Fg = -mg
Fd  Fg  0

This velocity is called the
terminal velocity.
cvt2 - mg  0
vt 
mg
c
Downhill Skiing


CBS Sports has invited you to be the special science
commentator for the Winter Olympics downhill ski
race.
You observe the following:
•
•
•
•
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The downhill course is 2.5 km long with a drop of 800 m
A downhill skier needs at least 3° of slope to move forward
The speed gun clocks the skier at a maximum of 130 km/h
An average skier is about 90 kg
What is the coefficient of kinetic friction and drag
coefficient for a skier?
Force Diagram
FN = mg cosq

At constant velocity the
forces must all balance.

Friction doesn’t act in the
direction of the normal force.

The normal force cancels the
component of gravity.
Fd = -cv2
Ffr = -mkFN
Fg = -mg
q
Fgy = -mg cosq
FN  mg cosq
Minimum Sliding
FN = mg cosq
Ffr = -mkFN
Fgx = mg sinq

When the skier is sliding to a
halt we can neglect drag.

Kinetic friction balances the
component of gravity pulling
forward.
Fgx  F fr  0
q  3°
mg sin q  m k mg cos q
m k  tan q  0.052
Downhill Run

FN = mg cosq
Fgx = mg sinq
• sinq = 800 m / 2500 m
 q = 19°
Fd = -cv2
Ffr = -mkFN
The slope of the downhill
course is

Drag force balances the
force of gravity and kinetic
friction.
Fgx  F fr  Fd  0
q  19°
mg sin q - m k mg cos q - cv 2  0
mg sin q - m k mg cos q
v2
c  0.19 kg/m
c