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AP Physics C: Mechanics : Friction and Normal Force

Study concepts, example questions & explanations for AP Physics C: Mechanics

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All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Forces

Weightlessness is experienced when the normal force equals __________.

Possible Answers:

the force due to gravity

zero

a negative value

the mass of the object

Correct answer:

zero

Explanation:

"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

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Example Question #2 : Forces

A book of mass 1kg is held to a vertical wall by a person's hand applying a 20N force directly toward the wall. The wall has a static friction coefficient of 0.3 and a kinetic friction coefficient of 0.2. With the book held at rest, what the is frictional force keeping the book from sliding down the wall?

Possible Answers:

39.2N

58.8N

Correct answer:

Explanation:

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

$F_f=\mu_sF_N$

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

F_f=(0.3)(20N)=6N

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Example Question #1 : Forces

$\begin{align*}&\text{A block of mass }190kg\text{ rests on an incline of }33^{\circ}\\&\text{It is also partially supported by a rope with a tension of }559.17N\\&\text{What is the normal force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

1563.2N

1094.24N

710.61N

1015.15N

Correct answer:

1094.24N

Explanation:

Normalwrope

$\begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\\&F_{normal}=1094.24N\end{align*}$

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Example Question #3 : Forces

$\begin{align*}&\text{A block of mass }233kg\text{ rests on an incline of }81^{\circ}\\&\text{What is the normal force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

230.13N

2257.59N

2285.73N

357.57N

Correct answer:

357.57N

Explanation:

Normal

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Example Question #3 : Forces

$\begin{align*}&\text{A block of mass }187kg\text{ rests on an incline of }83^{\circ}\\&\text{What is the normal force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

185.61N

223.57N

1820.8N

1834.47N

Correct answer:

223.57N

Explanation:

Normal

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Example Question #4 : Forces

$\begin{align*}&\text{A block of mass }267kg\text{ rests on an incline of }49^{\circ}\\&\text{What is the normal force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

1976.79N

2619.27N

201.51N

1718.4N

Correct answer:

1718.4N

Explanation:

Normal

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Example Question #5 : Forces

$\begin{align*}&\text{A block of mass }112kg\text{ rests on an incline of }18^{\circ}\\&\text{It is also partially supported by a rope.}\\&\text{The normal force acting on the block is }417.98N\\&\text{What is the tension felt in the rope?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

659.23N

1098.72N

1044.94N

339.52N

Correct answer:

659.23N

Explanation:

Normalwrope

$\begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{rope}=F_{mass}-\frac{F_{normal}}{cos(a)}\\&F_{rope}=1098.72N-\frac{417.98N}{cos(18^{\circ})}\\&F_{rope}=659.23N\end{align*}$

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Example Question #3 : Forces

A $\small 3500kg$ truck traveling at $\smlal 25\frac{m}{s}$ suddenly brakes and skids to a halt. If the coefficient of kinetic friction between the tires and the road is $\small 0.4$ , how far does the truck skid before stopping?

Possible Answers:

80m

24m

52m

Correct answer:

80m

Explanation:

Relevant equations:

$\Delta K = W$

$K = \frac{1}{2}mv^2$

W = F*d

$F_f = \mu F_N$

Determine the truck's change in kinetic energy. Note that the final kinetic energy will be zero because it has no final velocity.

$K_i = \frac{1}{2}(3500kg)(25\frac{m}{s})^2=1,093,750J$

K_f=0

$\Delta K = K_f - K_i = -1,093,750 J$

Using the work-energy theorem, set change in kinetic energy equal to the work done by friction.

$\Delta K = -F_f d$

Substitute the equation for force of friction.

$\Delta K=-\mu F_Nd$

$\Delta K = -\mu (mg)d$

Use the given values for coefficient of friction, mass, and acceleration of gravity to solve for the distance.

$-1,093,750 J= -(0.4)(3500kg)(9.8\frac{m}{s^2})d$

$d = 79.7m \approx 80m$

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Example Question #81 : Ap Physics C

A force of 40N is applied on a 14kg box to move it to the right. If the coefficient of friction between the box and the floor is 0.2, what is the acceleration of the box?

Possible Answers:

$2.0\frac{m}{s^2}$

$3.4\frac{m}{s^2}$

$0.2\frac{m}{s^2}$

$0.9\frac{m}{s^2}$

$0\frac{m}{s^2}$

Correct answer:

$0.9\frac{m}{s^2}$

Explanation:

The two horizontal forces acting on the box are the applied force of 40N to the right, and the friction force ( $\mu_kF_N$ )to the left. So the net horizontal force is written as the equation below.

$40N-\mu_kF_N=ma$

F_N=mg is the normal force. Solving for acceleration, a, we get the equation below.

$a=\frac{40N-\mu_kF_N}{m}$

We know the following from the question.

m=14kg

$g=9.8\frac{m}{s^2}$

$\mu_k=0.2$

$F_N=mg=(14kg)(9.8\frac{m}{s^2})=137.2N$

Thus, the acceleration is

$a=\frac{40N-(0.2)(137.2N)}{14kg}$

$a=0.9\frac{m}{s^2}$

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Example Question #4 : Forces

A $\small 15kg$ crate is pulled across a horizontal floor at a constant velocity of $\small 3\frac{m}{s}$ . The coefficient of kinetic friction between the crate and the floor is $\small 0.20$ . If the rope pulling the crate is angled at $\small 45^o$ above the horizontal, what force must be applied to maintain this constant velocity?

Possible Answers:

52N

62N

35N

48N

83N

Correct answer:

35N

Explanation:

Relevant equations:

$F_{net}= ma$

$F_f=\mu F_N$

F_g = mg

Write net force equations for horizontal and vertical directions:

Horizontal: $F_{net, x} = F_{pull, x}-F_f$

Vertical: $F_{net,y}=F_N+F_{pull,y}-F_g$

Since the crate is not accelerating, $\small a=0\frac{m}{s^2}$ in each direction, implying that $\small F_{net,x}=F_{net,y}=0N$ .

Horizontal: $0 = F_{pull,x}-F_{f}$

Vertical: $0 = F_N + F_{pull,y}-F_g$

Express the horizontal and vertical components of the pulling force in terms of the total pulling force, $F_{pull}$ .

Horizontal: $0 = (F_{pull}*cos(45^o))-F_f$

Vertical: $0 = F_N + (F_{pull}*sin(45^o))-F_g$

Replace force of friction with its expression in terms of the normal force.

Horizontal: $0 = (F_{pull}*cos(45^o))-(\mu F_N)$

Rearrange to isolate the normal force.

Horizontal: $F_N =\frac{F_{pull}*cos(45^o)}{\mu}$

Substitute this term for the normal force in the net vertical force equation.

Vertical: $0 = (\frac{F_{pull}*cos(45^o)}{\mu})+(F_{pull}*sin(45^o))-F_g$

Solve to isolate the pulling force.

$F_{pull}=\frac{F_g*\mu}{cos(45^o)+\mu*sin(45^o)}$

Use the given values for the mass of the box, coefficient of friction, and acceleration of gravity to solve for the pulling force.

$F_{pull}=\frac{(15kg)(9.8 \frac{m}{s^2})(0.20)}{cos(45^o)+0.20*sin(45^o)}$

$F_{pull}=\frac{29.4N}{0.849}=34.65N\approx35N$

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AP Physics C: Mechanics : Friction and Normal Force

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Forces

Example Question #2 : Forces

Example Question #1 : Forces

Example Question #3 : Forces

Example Question #3 : Forces

Example Question #4 : Forces

Example Question #5 : Forces

Example Question #3 : Forces

Example Question #81 : Ap Physics C

Example Question #4 : Forces

All AP Physics C: Mechanics Resources

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