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Online Quiz -
Friction
8B66-D878 |
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AP Physics Friction
Practice Test |
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1 |
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A block of mass = 2M is stacked atop a block of mass = M. The sliding COF
between the blocks = 0.8 and between the lower block and floor is 0.1. A
force = F is applied to the top block parallel to the ground. Find the
acceleration of the lower block. a = 1.3g |
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2 |
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Find the acceleration of a box of mangos sliding down a slope of 40 degrees.
Assume the dynamic COF = .2 and the static COF = .3. Would doubling the
number of mangos in the box have any effect on the acceleration?
No, a = g ( sinβ -
μd cosβ
). Note that mass is not in the equation. |
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3 |
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If the above box of mangos rests on a 20 degree slope, will it slide?
- qcr
= tan-1(0.3) or 16.7 degrees. It will slide.
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4 |
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Find the static COF if the box of mangos starts to slide if the angle of the
slope = 30 degrees. static COF = tan 30º
or 0.577 |
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5 |
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The tires on Martha's 4WD SUV have a static COF = 0.7 and a dynamic
COF = 0.5 with the road surface. What is her maximum acceleration and
minimum stopping distance given a starting velocity of 40 m/s? amax =
0.7g (Note that the maximum
acceleration comes from the max friction force, which is
the max static friction. With anti-lock brakes the wheels do not lock up and
slide. They roll slower and slower as the vehicle stops, at which time they
stop rolling.), Min stopping distance =
114 m |
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6 |
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If Martha's SUV is only 2WD, what is the effect on her maximum
acceleration (speeding up) and minimum stopping distance (slowing down)? Assume the COF's the same
for all wheels
and that the car's mass is evenly distributed (half over the front tires and
half over the back tires). Her maximum acceleration
(speeding up) is cut in half and her minimum stopping distance doubled.
Note that in reality there is a separate friction force acting on each
wheel. If the weight of the vehicle is evenly distributed, each driven wheel
will have a normal force of 1/4 mg making its static friction force = 1/4
μsmg.
4WD total friction = 1/4
μsmg
+1/4
μsmg+1/4
μsmg)+1/4
μsmg
= μsmg
2WD total friction = 1/4
μsmg
+1/4
μsmg
= 1/2 μsmg
Using kinematics analysis with amax
based on the maximum static friction:
( min stopping dist ) = vo
2 / (2amax)
Obviously, cutting the stopping
acceleration in half doubles the stopping distance.
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7 |
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Juan pushes downward on a box of mass = M with a force = 2Mg
that makes a 45 degree angle with the horizon. By contrast Susan pulls
upward on a similar box with a similar force and angle. Both boxes sit on a
surface with a sliding COF = 0.2 and a static COF = 0.3. Find the
acceleration of both boxes.
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Juan |
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Susan |
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Fn
= Mg
+ 2Mg sin 45º
= Mg(
1
+ 2 sin
45º )
Fd =
μd
Fn
= μd
Mg( 1 +
2 sin 45º
)
All of the acceleration is in the
horizontal dimension such that:
a = Σ F / m
= (
Fx
- Fd ) / M
= [2Mg cos
45º
-
μdMg( 1 + 2 sin
45º
) ] / M
=
2g cos
45º
-
μdg( 1 + 2 sin
45º
)
| a
= g ( 2cos
45º
-
μd
- 2μd sin
45º
) |
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Fn
= Mg
- 2Mg sin 45º
= Mg(
1
- 2 sin
45º
)
Fd
=
μd
Fn
= μd
Mg( 1 -
2 sin 45º
)
All of the acceleration is in the
horizontal dimension such that:
a = Σ F / m
= (
Fx
- Fd ) / M
= [2Mg cos
45º
-
μdMg( 1 - 2 sin
45º
) ] / M
=
2g cos
45º
-
μdg( 1 - 2 sin
45º )
| a
= g ( 2cos
45º
-
μd
+ 2μd sin
45º ) |
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