Download Example 5-2 Sliding with Kinetic Friction

Example 5-2 Sliding with Kinetic Friction
In Example 4-6 (see Section 4-6), we considered a book sliding down an incline with friction. In a variation of this problem, suppose the coefficient of kinetic friction between the incline and the sliding 2.50-kg book is 0.350. (a) What is the
downhill acceleration of the book if the incline is at an angle from the horizontal of 30.0°? (b) What is the acceleration if
the angle is 10.0°?
Set Up
The book is moving down the incline of
angle u, so the kinetic friction force (which
always opposes sliding) must point up the
incline. To find the net external force on the
book and hence the book’s acceleration, we
need to determine the magnitude of the friction
force by using Equation 5-6. Once we’ve found
a general expression for the block’s acceleration
in terms of u, we’ll substitute the values u = 30.0°
and u = 10.0°.
Solve
Write Newton’s second law for the block
in component form. Because the block’s
acceleration is along the incline (that is, along
the x direction), it follows that ablock,y = 0.
n
Newton’s second law applied to
the book:
s
s + w
s book + sf k
a Fext on block = n
= mblocks
ablock
motion
of book
Magnitude of the kinetic friction
force:
fk = mkn
fk
y
x
wbook
O
(5-6)
Newton’s second law for the book
in component form:
x: a Fext on book, x
= 0 + wbook sin u + (2fk)
= mbook abook, x
y: a Fext on book, y
= n 2 wbook cos u + 0
= mbook abook,y = 0
ny = n
f k,x = –f k
y
x
wbook
wbook,y = –wbook cos O
O
wbook,x = wbook sin O
To find the magnitude of the kinetic friction
force, we solve the y component equation for
the magnitude of the normal force n and then
substitute this expression into the equation for
fk. Next we substitute our result for fk into the
x component equation, giving an equation for
the acceleration abook,x.
From the y component equation,
n = wbook cos u
Substitute this into Equation 5-6:
fk = mkn = mk(wbook cos u)
Substitute this expression for fk into the x equation:
wbook sin u 2 mkwbook cos u = mbookabook, x
Now we can complete our solution for the
book’s acceleration abook, x. Use the relationship
between gravitational force and mass.
Relationship between the mass mbook of the book and the gravitational force wbook on the book:
wbook = mbook g
Substitute into the x equation and solve for abook, x:
mbook g sin u 2 mk mbook g cos u = mbook abook, x
The factor of mbook cancels, so
abook, x = g sin u 2 mk g cos u or
abook, x = g(sin u 2 mk cos u)
Use this general formula for abook, x for the
given value of mk and the two given values of u.
(a) Substitute u = 30.0°:
abook,x = g(sin u 2 mk cos u)
= 19.80 m>s 2 2 3 sin 30.0 - 10.3502 1cos 30.02 4
= 19.80 m>s 2 2 3 0.500 - 10.3502 10.8662 4
= 1.93 m>s 2
(b) Substitute u = 10.0°:
abook, x = g(sin u 2 mk cos u)
= 19.80 m>s 2 2 3 sin 10.0 - 10.3502 1cos 10.02 4
= 19.80 m>s 2 2 3 0.174 - 10.3502 10.9852 4
= -1.68 m>s 2
Reflect
The book has a positive (downhill) acceleration
for u = 30.0°, which means the book speeds up
as it slides downhill. However, when the incline
is at a shallower angle of u = 10.0°, the book
Downhill component of gravitational force:
wbook, x = wbook sin u = mbook g sin u
Uphill component of kinetic friction force:
fk = mkn = mkmbook g cos u
has a negative (uphill) acceleration: Its speed
decreases as it moves down the incline.
To see why the speed decreases, let’s
calculate the downhill (x) component of the
gravitational force and the uphill component
of the kinetic friction force for each angle. The
shallower the angle u, the smaller the downhill
gravitational force and the greater the uphill
force of kinetic friction. If the angle is small
enough, the friction force will dominate over
the gravitational force—in which case the net
force and acceleration point up the incline.
Can you show that the acceleration of the
block will be zero (that is, it will slide down the
incline with constant velocity) if u = 19.3°?
If u = 30.0°,
wbook,x = 12.50 kg2 19.80 m>s 2 2 sin 30.0
= 12.3 N
fk = 10.3502 12.50 kg2 19.80 m>s 2 2 cos 30.0
= 7.43 N
so gravity dominates over kinetic friction and the net force is
downhill.
If u = 10.0°,
wbook,x = 12.50 kg2 19.80 m>s 2 2 sin 10.0
= 4.25 N
fk = 10.3502 12.50 kg2 19.80 m>s 2 2 cos 10.0
= 8.44 N
Kinetic friction dominates over gravity and so the net force is uphill.