Friction - AP Physics C: Electricity and Magnetism

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Question

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?

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Answer

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.

$40N-\mu_kF_N=ma$

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.

$\mu_k=0.2$

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

Thus, the acceleration is

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

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Question

A truck driver is making a delivery. He takes a 287kg box out of his semi-truck by having it slide down a 33o ramp that is 5.5m long. To ensure that the box doesn't accelerate down the ramp, the driver pushes back on it so that the box's constant speed is $1.7$\frac{m}{s}$$ . If the coefficient of friction is 0.4, calculate the work done by the frictional force.

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Answer

We can calculate the work done by friction by using .

It is just the product of the friction force and the length of the ramp. Rewrite work as $W=\mu_kF_Nd$ .

Since this problem involves a box sliding down an incline, the normal force is $F_N=mg\cos\theta$ .

Work is now written as the equation below.

$W=\mu_kmg(\cos\theta) d$

We know the following information from the question.

$\mu_k=0.4$

using these values, we can calculate work from our equation.

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Question

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?

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Answer

Relevant equations:

$F_f=\mu F_N$

Write net force equations for horizontal and vertical directions:

Horizontal:

Vertical:

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, .

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.

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

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Question

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?

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Answer

Relevant equations:

$\Delta K = W$

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

$F_f = \mu F_N$

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$

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

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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Question

A string connected to a box sitting on the floor is used to apply a
force on the box as shown. The string is inclined at an angle of 30°
above the horizontal, the box has a mass of 15 kg, and the tension
in the string is 40 N. The box is at rest, but on the verge of slipping.
What is the coefficient of static friction between the box and the
floor?

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Answer

First, we know that the tension on the string is at an angle of degrees from the horizontal. We must find the x and y components of the tension force.

$T_y = T\sin(30) = 40\cdot\sin(30)= 20:N$

$T_x = T\cos(30) = 40\cos(30)= 34.64:N$

Next, we must find the normal force acting on the block. The normal force of the block itself is $15:kg \cdot $9.8:m/s^2$ = 147:N$ , but there is a string pulling on it. We must subtract the y component of tension to get the final normal force

Next, we know that the block is being pulled to the right, and the frictional force is acting in the opposite direction. So we get the equation:

$T_x - F_f = 0, F_f = \mu F_n$

$34.64 - \mu 127 = 0$

$\mu = 0.273$

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Question

$\begin{align}&$\text{A block of mass }$100kg$\text{ rests on an incline of $}$73^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a)\&F_${friction}=100kg(9.81$\frac{m}{s^2$$}$)(sin(73^${\circ}$))\&F_{friction}$=938.13N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$51kg$\text{ rests on an incline of $}$11^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a))\&F_${friction}=51kg(9.81$\frac{m}{s^2$$}$)(sin(11^${\circ}$)\&F_{friction}$=95.46N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$19kg$\text{ rests on an incline of $}$69^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a)\&F_${friction}=19kg(9.81$\frac{m}{s^2$$}$)(sin(69^${\circ}$))\&F_{friction}$=174.01N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$284kg$\text{ rests on an incline of $}$48^{\circ}$\&$\text{It is also partially supported by a rope.}$\&$\text{The frictional force acting on the block is }$207.04N\&$\text{What is the tension felt in the $rope?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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}$$=mg\&F_{normal}$$=(F_{mass}$$-F_{rope}$$)cos(a)\&F_{friction}$$=(F_{mass}$$-F_{rope}$$)sin(a)\&F_{rope}$$=F_{mass}$$-$\frac{F_{friction}$$}{sin(a)}\&F_${rope}=2786.04N-$\frac{207.04N}{sin(48^${\circ}$$)}\&F_{rope}$=2507.44N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$101kg$\text{ rests on an incline of $}$68^{\circ}$\&$\text{It is also partially supported by a rope with a tension of }$693.57N\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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}$$=mg\&F_{normal}$$=(F_{mass}$$-F_{rope}$$)cos(a)\&F_{friction}$$=(F_{mass}$$-F_{rope}$)sin(a)\&F_${friction}=(101kg(9.81$\frac{m}{s^2$$}$)-693.57N)(sin(68^${\circ}$))\&F_{friction}$=275.6N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$187kg$\text{ rests on an incline of $}$70^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a)\&F_${friction}=187kg(9.81$\frac{m}{s^2$$}$)(sin(70^${\circ}$))\&F_{friction}$=1723.84N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$208kg$\text{ rests on an incline of $}$28^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

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Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a)\&F_${friction}=208kg(9.81$\frac{m}{s^2$$}$)(sin(28^${\circ}$))\&F_{friction}$=957.95N\end{align*}$

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Question

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?

Tap to reveal answer

Answer

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.

$40N-\mu_kF_N=ma$

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.

$\mu_k=0.2$

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

Thus, the acceleration is

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

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Question

A truck driver is making a delivery. He takes a 287kg box out of his semi-truck by having it slide down a 33o ramp that is 5.5m long. To ensure that the box doesn't accelerate down the ramp, the driver pushes back on it so that the box's constant speed is $1.7$\frac{m}{s}$$ . If the coefficient of friction is 0.4, calculate the work done by the frictional force.

Tap to reveal answer

Answer

We can calculate the work done by friction by using .

It is just the product of the friction force and the length of the ramp. Rewrite work as $W=\mu_kF_Nd$ .

Since this problem involves a box sliding down an incline, the normal force is $F_N=mg\cos\theta$ .

Work is now written as the equation below.

$W=\mu_kmg(\cos\theta) d$

We know the following information from the question.

$\mu_k=0.4$

using these values, we can calculate work from our equation.

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Question

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?

Tap to reveal answer

Answer

Relevant equations:

$F_f=\mu F_N$

Write net force equations for horizontal and vertical directions:

Horizontal:

Vertical:

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, .

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.

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

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Question

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?

Tap to reveal answer

Answer

Relevant equations:

$\Delta K = W$

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

$F_f = \mu F_N$

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$

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

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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Question

A string connected to a box sitting on the floor is used to apply a
force on the box as shown. The string is inclined at an angle of 30°
above the horizontal, the box has a mass of 15 kg, and the tension
in the string is 40 N. The box is at rest, but on the verge of slipping.
What is the coefficient of static friction between the box and the
floor?

Tap to reveal answer

Answer

First, we know that the tension on the string is at an angle of degrees from the horizontal. We must find the x and y components of the tension force.

$T_y = T\sin(30) = 40\cdot\sin(30)= 20:N$

$T_x = T\cos(30) = 40\cos(30)= 34.64:N$

Next, we must find the normal force acting on the block. The normal force of the block itself is $15:kg \cdot $9.8:m/s^2$ = 147:N$ , but there is a string pulling on it. We must subtract the y component of tension to get the final normal force

Next, we know that the block is being pulled to the right, and the frictional force is acting in the opposite direction. So we get the equation:

$T_x - F_f = 0, F_f = \mu F_n$

$34.64 - \mu 127 = 0$

$\mu = 0.273$

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Question

$\begin{align}&$\text{A block of mass }$100kg$\text{ rests on an incline of $}$73^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

Tap to reveal answer

Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a)\&F_${friction}=100kg(9.81$\frac{m}{s^2$$}$)(sin(73^${\circ}$))\&F_{friction}$=938.13N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$51kg$\text{ rests on an incline of $}$11^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

Tap to reveal answer

Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a))\&F_${friction}=51kg(9.81$\frac{m}{s^2$$}$)(sin(11^${\circ}$)\&F_{friction}$=95.46N\end{align*}$

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Question

$\begin{align}&$\text{A block of mass }$19kg$\text{ rests on an incline of $}$69^{\circ}$\&$\text{What is the frictional force acting on the block from the $surface?}$\&g=9.81$\frac{m}{s^2$}$\end{align}$

Tap to reveal answer

Answer

$\begin{align}&$\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), a normal force, and since it’s on an incline, a frictional}$\&$\text{force to balance it all $out:}$\&F_{mass}$$=mg\&F_{normal}$$=mgcos(a)\&F_{friction}$=mgsin(a)\&F_${friction}=19kg(9.81$\frac{m}{s^2$$}$)(sin(69^${\circ}$))\&F_{friction}$=174.01N\end{align*}$

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