Practice Test: Mr Rogers AP Physics C -- Friction

 Unit Plan Practice Test Study Guide

Online Quiz - Friction                                                                  8B66-D878

AP Physics Friction Practice Test

1 ) 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

2 ) 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.

3 )
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.
4 ) 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

5 ) 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

6 ) 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.

7 ) 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.
Juan Susan

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 )

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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