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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #21 : Force Of Friction

How will changing the amount of surface area in contact between two surfaces affect friction?

Possible Answers:

Increasing surface area will increase friction to a point, then decrease it

Increasing surface area will decrease friction

It depends on the surfaces involved

Increasing surface area will increase friction

It is unaffected

Correct answer:

It is unaffected

Explanation:

Frictional force is independent of surface area, it only depends on the normal force and the coefficient of friction.

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Example Question #732 : Ap Physics 1

A $100\textup{ kg}$ box is in a moving bus. When the bus is moving at $30\ \frac{\textup{m}}{\textup{s}}$ and goes around a corner with a radius of $20\textup{ m}$ , the force is just enough to cause the box to “move to the outside” of the bus. Determine the coefficient of static friction between the floor and the box.

Possible Answers:

4.59

8.44

6.95

None of these

7.47

Correct answer:

4.59

Explanation:

Right when the box starts to slide, the magnitude of the frictional force is equal to the magnitude of the centripetal force.

$F_{friction}=m\frac{v^2}{r}$

Using the definition of frictional force.

$\mu*g*m=m\frac{v^2}{r}$

Solving for $\mu$

$\mu=\frac{v^2}{rg}$

Plugging in values:

$\mu=\frac{30^2}{20*9.8}$

$\mu=4.59$

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Example Question #21 : Force Of Friction

What coefficient of friction would be needed to stop a 150kg box sliding across a level surface at $3\frac{m}{s}$ in ?

Possible Answers:

None of these

Correct answer:

Explanation:

E_i+W=E_f

$.5mv_{i}^2-F_{friction}*d=.5mv_{f}^2$

$.5mv_{i}^2-\mu*mg*d=.5mv_{f}^2$

Solving for $\mu$ :

$\mu=.5\frac{v_i^2-v_f^2}{g*d}$

Plugging in values:

$\mu=.5\frac{3^2-0^2}{9.8*5}$

$\mu=.092$

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Example Question #21 : Force Of Friction

Ramp force diagram

A $90\textup{ kg}$ box starts sliding down an adjustable ramp when the angle has reached $19^\circ$ . Determine the coefficient of friction between the box and the ramp.

Possible Answers:

None of these

Correct answer:

Explanation:

By constructing a force diagram, it can be seen that:

$F_{friction}=\mu*N=\mu*m*g*cos(\theta)$

$F_{gravity down ramp}=mgsin(\theta)$

Right when the box starts to move, these will be equal

$m*g*sin(\theta)=\mu*m*g*cos(\theta)$

Solving for $\mu$

$\mu=\frac{sin(\theta)}{cos(\theta)}$

Plugging in values:

$\mu=.344$

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Example Question #22 : Force Of Friction

Pulley with friction

2 masses are connected by a pulley as demonstrated by the image above. The weight of mass 1 is $19\ kg$ and and the weight of mass 2 is $21\ kg$ . The coefficient of static friction between mass 1 and the block it rests on is $\mu _{s}=0.96$ and the coefficient of kinetic friction between the 2 surfaces is $\mu _{k}=.65$ . The system is not drawn to scale.

How fast will mass 1 be accelerating in this system?

Possible Answers:

$3.6\ \frac{m}{s^{2}}$

$6.8\ \frac{m}{s^{2}}$

$2.2\ \frac{m}{s^{2}}$

$9.1\ \frac{m}{s^{2}}$

$4.5\ \frac{m}{s^{2}}$

Correct answer:

$4.5\ \frac{m}{s^{2}}$

Explanation:

Because the mass of object 1 is given, the force acting on it must be found in order to figure out how fast it is accelerating. These 3 variables are related by the equation:

F=ma

There are 2 forces acting on mass 1. The first force is the tension in the rope. This tension is equal to the magnitude of the gravitational force created by object 2. This can be found using the equation:

F=mg

where g is the gravity constant which is equal to $9.81\frac{m}{s^{2}}$ .

The other force acting on the object is friction. The force of friction can be found by multiplying the normal force exerted by the object on the surface that it is resting on by the coefficient of friction. This is demonstrated by the equation:

$F=\mu*N$

Where $\mu$ is the coefficient of friction and N is the normal force. Static friction is the force of friction acting on an object that is not moving. because the magnitude of the force of static friction would not be as great as the magnitude of the force of tension in the rope the object would be moving. Because the object is moving, kinetic friction would be the other force that is acting on mass 1. Kinetic friction represents the friction exerted on an object that is moving on a surface.

Because there are 2 forces acting on this object, a free body diagram can be made to illustrate these forces.

Fbd

T represents tension in the rope while F represents friction. The total net force can be found by subtracting the force of friction from the magnitude of tension in the rope.

$F=m_{2}g-\mu _{k}N$

F=84.86N

Acceleration of mass 1 can now be figured out because the force acting on the object has been found and the mass is given.

$84.86=19a_{1}$

$a_{1}=4.5\ \frac{m}{s^{2}}$

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Example Question #121 : Forces

Pulley with friction

Mass 1 weighs $26\ kg$ and mass 2 weighs $19\ kg$ .

What would be the minimum coefficient of static friction between mass 1 and the surface that would cause mass 1 to not slide in this system?

Possible Answers:

1.2

0.73

0.23

0.55

0.84

Correct answer:

0.73

Explanation:

In order for mass 1 to remain stationary while mass 2 is experiencing free fall, the magnitude of the force of static friction acting on mass 1 must be greater than the magnitude of the force of tension in the rope that is created by mass 2. The force of tension in the rope is equal to the gravitational force that is created by mass 2. This force can be found using the equation:

F=mg

The minimum coefficient of static friction would create a force in the opposite direction that has the same magnitude as the force found above. Force of static friction is equal to the coefficient static friction between an object and surface multiplied by the normal force exerted by the object being analyzed. This is represented by the following equation:

$F=\mu _{s}N$

Where $\mu _{s}$ is the coefficient of static friction and N is normal force. Normal force in this problem would be equal to the mass of object 1 multiplied by the gravity constant. By setting the 2 equations above equal to each other, minimum coefficient of static friction can be found.

$m_{2}g=\mu _{s}N$

After plugging in given values,

$19*9.81=26(9.81)\mu _{s}$

$\mu _{s}=.73$

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Example Question #24 : Force Of Friction

How will doubling the coefficient of friction of rubber tires change the stopping distance of a car?

Possible Answers:

Impossible to determine

Cut the stopping distance in half

None of these

Quadruple the stopping distance

Double the stopping distance

Correct answer:

Cut the stopping distance in half

Explanation:

W=KE_f-KE_i

W=.5mv_f^2-.5mv_i^2

$W=F_{friction}*d$

$F_{friction}=\mu*m*g$

$W=\mu*m*g*d$

Since the car is stopped at the end, v_f=0

Combining equations:

$.5mv_i^2=\mu*m*g*d$

$.5v_i^2=\mu*g*d$

Solving for

$\frac{.5v_i^2}{\mu*g}=d$

From this, it can be seen that doubling the coefficient of friction would cut the stopping distance in half.

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Example Question #731 : Ap Physics 1

A box is shoved and is sliding on a concrete floor. It has a mass of 50kg and in 2.0meters slows from $5\frac{m}{s}$ to $3\frac{m}{s}$ . Determine the coefficient of friction.

Possible Answers:

None of these

Correct answer:

Explanation:

W=E_f-E_i

W=.5*50*3^2-.5*50*5^2

W=-400J

W=F*d

Work is due to friction:

$W=\mu*m*g*d$

Solving for $\mu$

$\mu=\frac{W}{mgd}$

Plugging in values:

$\mu=\frac{400}{50*9.8*2}$

$\mu=.41$

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Example Question #731 : Ap Physics 1

A box is shoved and is sliding on a concrete floor. It has a mass of 150kg and in 3.5m slows from $6\frac{m}{s}$ to $1\frac{m}{s}$ . Determine the coefficient of friction.

Possible Answers:

None of these

Correct answer:

Explanation:

W=E_f-E_i

W=.5*150*1^2-.5*150*6^2

W=-2663J

W=F*d

Work is due to friction:

$W=\mu*m*g*d$

Solving for $\mu$

$\mu=\frac{W}{mgd}$

Plugging in values:

$\mu=\frac{-2663}{150*-9.8*3.5}$

$\mu=.51$

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Example Question #701 : Newtonian Mechanics

A 5kg block is held at rest at the top of slope inclined at $20\degree$ . When the block is let go it slides down the slope, experiencing friction along the way. If the acceleration of the block down the ramp is $a=-5\frac{m}{s^2}$ , what is the coefficient of kinetic friction between the block and ramp?

Possible Answers:

1.12

0.89

1.43

0.99

0.56

Correct answer:

1.43

Explanation:

Because the block is traveling on an inclined ramp and experiences friction, the two main forces at work here are a component of gravitational force and kinetic frictional force.

The total force acting on the block can be written as $F_{total} = F_{g, \parallel } + f_{k}$ , where $F_{g, \parallel }$ is the component of gravitational force in the direction of motion, and $f_{k}$ is the kinetic frictional force.

We know that forces can be expressed as F=ma , and kinetic friction can be expressed as $f_{k} = -\mu_k N$ . Hence our total force can be expressed as $F_{total} = ma_{total} = ma_{g, \parallel } - \mu_{k}N$ . The normal force can be further expressed as $N = ma_{g, \perp }$

The component of gravity parallel to the ramp is given by $a_{g,\parallel} = a_{total}sin(20\degree)$ , while the component of gravity perpendicular to the ramp is given by $a_{g, \perp }=a_{total}cos(20\degree)$ .

Our total force can hence be expressed as $ma_{total} = ma_{g, \parallel } - \mu_{k}ma_{g, \perp }$ .

Thus, $\mu_k = \frac{ma_{g, \parallel } - ma_{total}}{ma_{g, \perp }} = \frac{a_{total}sin(20 \degree) - a_{total}}{a_{total}cos(20\degree)}$

and $\mu_k = \frac{(-5\frac{m}{s^2})sin(20\degree) - (-5 \frac{m}{s^2})}{(-5\frac{m}{s^2})cos(20\degree)} = 1.43$

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #21 : Force Of Friction

Example Question #732 : Ap Physics 1

Example Question #21 : Force Of Friction

Example Question #21 : Force Of Friction

Example Question #22 : Force Of Friction

Example Question #121 : Forces

Example Question #24 : Force Of Friction

Example Question #731 : Ap Physics 1

Example Question #731 : Ap Physics 1

Example Question #701 : Newtonian Mechanics

All AP Physics 1 Resources

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