Kinetic and Static Friction - AP Physics C Mechanics

A $10.0\,\text{kg}$ sled slides on packed snow (sled–snow contact) with $\mu_s=0.20$ and $\mu_k=0.10$ . A student pulls horizontally with $30\,\text{N}$ and the sled is already moving. Based on the scenario, calculate the force of kinetic friction acting on the sled.
Answer: $9.8\,\text{N}$ This question tests AP Physics C: Mechanics skills, specifically understanding of kinetic and static friction and their roles in force dynamics. Kinetic friction acts on moving objects with a constant magnitude equal to μk times the normal force, independent of the object's speed or applied force. In this scenario, a 10.0 kg sled is already moving on packed snow with μk = 0.10, and a student pulls with 30 N horizontally. Choice A (9.8 N) is correct because kinetic friction equals μk × N = μk × mg = 0.10 × 10.0 × 9.8 = 9.8 N, regardless of the 30 N applied force. Choice B (19.6 N) incorrectly uses the static coefficient or doubles the kinetic friction, while choice C uses the applied force magnitude. Students should understand that kinetic friction depends only on the coefficient and normal force, not on the applied force or velocity. Draw free-body diagrams showing all forces and emphasize that kinetic friction opposes motion with constant magnitude.
A car of mass $m$ travels at a constant speed $v$ around a flat, circular track of radius $R$ . The coefficient of static friction between the tires and the road is $\mu_s$ . What is the magnitude of the friction force acting on the car?
Answer: $m v^2 / R$ For the car to move in a circle, there must be a net force directed towards the center of the circle (a centripetal force). This force is provided by static friction between the tires and the road. The required centripetal force has a magnitude of $F_c = m a_c = m v^2 / R$ . Since static friction is the only horizontal force, its magnitude must be equal to this value. Note that this force must be less than or equal to the maximum static friction, $\mu_s mg$ , but it is not necessarily equal to it.
Consider a block of mass $M$ on a rough horizontal surface. It is acted upon by a horizontal force $F$ . If the static friction force acting on the block is $f_s$ , which of the following inequalities must be true if the block remains at rest?
Answer: $f_s = F$ If the block remains at rest, it is in static equilibrium, and the net force on it is zero. The applied horizontal force $F$ is balanced by the static friction force $f_s$ . Therefore, the magnitude of the static friction force must be equal to the magnitude of the applied force, $f_s = F$ . This is true as long as $F$ does not exceed the maximum static friction, $F \le \mu_s Mg$ . The static friction force adjusts its magnitude to match the applied force.
A person pushes a 50 kg crate horizontally with a constant force of 200 N on a level floor. The crate accelerates at 1.0 m/s². What is the coefficient of kinetic friction between the crate and the floor?
Answer: 0.31 First, apply Newton's second law in the horizontal direction: $F_{net} = F_{applied} - F_k = ma$ . The kinetic friction force is $F_k = F_{applied} - ma = 200 \, \text{N} - (50 \, \text{kg})(1.0 \, \text{m/s}^2) = 150 \, \text{N}$ . The magnitude of the kinetic friction force is also given by $F_k = \mu_k F_N$ . On a level floor, $F_N = mg = (50 \, \text{kg})(9.8 \, \text{m/s}^2) = 490 \, \text{N}$ . Therefore, $\mu_k = F_k / F_N = 150 \, \text{N} / 490 \, \text{N} \approx 0.306$ , which is closest to 0.31.
A 10 kg block rests on a horizontal surface. The coefficient of static friction between the block and the surface is 0.50, and the coefficient of kinetic friction is 0.40. A horizontal force is applied to the block, and its magnitude is slowly increased from zero. At what applied force does the block begin to move?
Answer: The block begins to move when the applied force just exceeds 49.0 N. The block begins to move when the applied horizontal force exceeds the maximum force of static friction, $F_{s,max} = \mu_s F_N$ . On a horizontal surface, the normal force $F_N$ is equal to the weight of the block, $mg$ . Using $g \approx 9.8 \, \text{m/s}^2$ , $F_N = (10 \, \text{kg})(9.8 \, \text{m/s}^2) = 98 \, \text{N}$ . Therefore, $F_{s,max} = (0.50)(98 \, \text{N}) = 49.0 \, \text{N}$ . The block will start to move when the applied force is just greater than 49.0 N.
A block of mass $m$ is pushed against a vertical wall by a horizontal force of magnitude $P$ . The coefficient of static friction between the block and the wall is $\mu_s$ . What is the minimum magnitude of $P$ required to prevent the block from sliding down the wall?
Answer: $mg / \mu_s$ For the block to be held in place, the upward static friction force must balance the downward gravitational force, $F_s = mg$ . The static friction force cannot exceed its maximum value, $F_s \leq \mu_s F_N$ . The normal force $F_N$ is provided by the wall and is equal in magnitude to the applied horizontal force $P$ , so $F_N = P$ . The condition for the block not to slide is $mg \leq \mu_s P$ . Therefore, the minimum force $P$ required is when $mg = \mu_s P$ , which gives $P = mg / \mu_s$ .
A 5.0 kg block is placed on a rough inclined plane that makes an angle of 30° with the horizontal. The block remains at rest. What is the magnitude of the static friction force acting on the block?
Answer: The static friction force is approximately 24.5 N, acting up the incline. Since the block is at rest, it is in static equilibrium. The net force on the block is zero. The forces acting parallel to the incline are the component of gravity pulling the block down the incline ( $mg \sin\theta$ ) and the static friction force ( $F_s$ ) acting up the incline. Therefore, $F_s = mg \sin\theta$ . Using $g \approx 9.8 \, \text{m/s}^2$ , $F_s = (5.0 \, \text{kg})(9.8 \, \text{m/s}^2)(\sin 30°) = (49 \, \text{N})(0.5) = 24.5 \, \text{N}$ . The static friction force adjusts its magnitude to be equal and opposite to the net force that would cause motion, as long as that force does not exceed the maximum static friction.
A heavy sled is pulled by a rope. The work required to pull the sled at a constant speed $v$ over a distance $d$ on a rough horizontal snowfield is $W$ . What is the work required to pull the same sled at a constant speed $2v$ over the same distance $d$ ? Assume the coefficient of kinetic friction is independent of speed.
Answer: $W$ The work done against friction is given by $W = F_{pull} \cdot d$ . Since the sled is pulled at a constant speed, the pulling force must be equal in magnitude to the kinetic friction force, $F_{pull} = F_k = \mu_k F_N$ . The coefficient of kinetic friction is assumed to be independent of speed, and the normal force is constant. Therefore, the friction force $F_k$ is the same at speed $v$ and speed $2v$ . Since the pulling force and the distance are the same in both cases, the work done is also the same, $W$ .
Which of the following statements about static and kinetic friction is always correct?
Answer: The coefficient of static friction is generally greater than or equal to the coefficient of kinetic friction. The coefficient of static friction ( $\mu_s$ ) is an empirical property that is typically found to be greater than or equal to the coefficient of kinetic friction ( $\mu_k$ ) for the same pair of surfaces. Choice A is incorrect because the static friction force can be any value from zero up to its maximum, which could be less than the kinetic friction force. Choice C is incorrect because the static friction force adjusts to oppose the applied force and is only equal to $\mu_s F_N$ at its maximum value, just before slipping. Choice D is not always true; friction is proportional to the normal force, which is not always equal to the object's weight (e.g., on an incline or with other vertical forces present).
A puck is given an initial speed $v_0$ and slides up a ramp inclined at an angle $\theta$ . The coefficient of kinetic friction between the puck and the ramp is $\mu_k$ . What is the magnitude of the puck's acceleration as it slides up the ramp?
Answer: $g(\sin\theta + \mu_k \cos\theta)$ When the puck slides up the ramp, two forces act on it parallel to the ramp's surface and directed down the ramp: the component of gravity $mg\sin\theta$ and the kinetic friction force $F_k$ . The normal force is $F_N = mg\cos\theta$ , so the friction force is $F_k = \mu_k F_N = \mu_k mg\cos\theta$ . The net force down the ramp is $F_{net} = mg\sin\theta + \mu_k mg\cos\theta$ . According to Newton's second law, $F_{net} = ma$ . Therefore, $ma = mg\sin\theta + \mu_k mg\cos\theta$ . Dividing by $m$ gives the acceleration $a = g(\sin\theta + \mu_k \cos\theta)$ .
A block of mass $m$ rests on a larger block of mass $M$ , which rests on a frictionless horizontal table. The coefficient of static friction between the blocks is $\mu_s$ . A horizontal force $F$ is applied to the lower block $M$ . What is the maximum force $F$ that can be applied such that the block $m$ does not slip relative to $M$ ?
Answer: $\mu_s (m+M)g$ The top block $m$ accelerates due to the static friction force from the bottom block, $F_s$ . The maximum acceleration of the top block is $a_{max} = F_{s,max}/m = \mu_s F_N / m = \mu_s mg / m = \mu_s g$ . For the top block not to slip, the entire system (both blocks) must have this as its maximum acceleration. Applying Newton's second law to the combined system of mass $(m+M)$ , the maximum applied force is $F_{max} = (m+M)a_{max} = (m+M)\mu_s g$ .
A block is projected with an initial velocity $v_0$ across a horizontal surface with a coefficient of kinetic friction $\mu_k$ . How much work is done by the friction force as the block slides to a stop?
Answer: $W_f = -\frac{1}{2}mv_0^2$ According to the work-energy theorem, the net work done on an object is equal to its change in kinetic energy: $W_{net} = \Delta K$ . In this case, the net work is done by the friction force, $W_{net} = W_f$ . The change in kinetic energy is $\Delta K = K_f - K_i$ . Since the block comes to a stop, $K_f = 0$ . The initial kinetic energy is $K_i = \frac{1}{2}mv_0^2$ . Therefore, $W_f = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$ . The work is negative because the friction force opposes the displacement.
Two blocks, A and B, are made of the same material and have the same mass. Block A has a square base with side length $L$ , and Block B has a rectangular base of length $2L$ and width $L/2$ . Both are placed on the same rough, horizontal surface. How does the maximum static friction force $F_{s,max,A}$ on Block A compare to the maximum static friction force $F_{s,max,B}$ on Block B?
Answer: $F_{s,max,A} = F_{s,max,B}$ The maximum static friction force is given by $F_{s,max} = \mu_s F_N$ . The area of the base of Block A is $L^2$ and for Block B is $(2L)(L/2) = L^2$ . Although the areas are the same, the friction force is independent of the contact area. Since both blocks have the same mass, their weight $mg$ is the same, and the normal force $F_N$ from the horizontal surface is the same for both. They are made of the same material and are on the same surface, so $\mu_s$ is the same. Therefore, the maximum static friction force is the same for both blocks.
A coin of mass $m$ is placed on a horizontal turntable at a distance $r$ from the center. The turntable rotates with a constant angular speed $\omega$ . The coefficient of static friction between the coin and the turntable is $\mu_s$ . What is the maximum angular speed $\omega_{max}$ at which the coin can rotate without slipping?
Answer: $(\mu_s g / r)^{1/2}$ The centripetal force required to keep the coin in a circular path is provided by the static friction force. The required centripetal force is $F_c = m a_c = m \omega^2 r$ . The maximum available static friction force is $F_{s,max} = \mu_s F_N = \mu_s mg$ . The coin will not slip as long as $m \omega^2 r \leq \mu_s mg$ . To find the maximum angular speed, we set these equal: $m \omega_{max}^2 r = \mu_s mg$ . Solving for $\omega_{max}$ gives $\omega_{max}^2 = \mu_s g / r$ , so $\omega_{max} = (\mu_s g / r)^{1/2}$ .
A student wants to determine the coefficient of static friction $\mu_s$ between a block and a wooden board. The student places the block on the board and slowly raises one end of the board, increasing the angle of inclination $\theta$ until the block just begins to slide. Which of the following expressions represents $\mu_s$ in terms of the angle $\theta_{max}$ at which the block starts to slide?
Answer: $\mu_s = \tan\theta_{max}$ At the point where the block is about to slide, the static friction force has reached its maximum value, $F_{s,max} = \mu_s F_N$ . This force must balance the component of gravity parallel to the incline, $mg \sin\theta_{max}$ . The normal force is $F_N = mg \cos\theta_{max}$ . Setting the forces parallel to the incline equal: $mg \sin\theta_{max} = \mu_s (mg \cos\theta_{max})$ . The term $mg$ cancels from both sides, leaving $\sin\theta_{max} = \mu_s \cos\theta_{max}$ . Solving for $\mu_s$ gives $\mu_s = \sin\theta_{max} / \cos\theta_{max} = \tan\theta_{max}$ .
A crate is at rest on a rough horizontal surface. A person can either push the crate with a force $P$ directed at an angle $\theta$ below the horizontal, or pull the crate with a force $P$ directed at the same angle $\theta$ above the horizontal. To make the crate start moving, which method requires a smaller force $P$ , and why?
Answer: Pulling, because the upward component of the pull reduces the normal force and thus the friction. To start the crate moving, the horizontal component of the applied force, $P\cos\theta$ , must overcome the maximum static friction, $\mu_s F_N$ . When pushing down at an angle, the normal force is $F_N = mg + P\sin\theta$ . When pulling up at an angle, the normal force is $F_N = mg - P\sin\theta$ . Since the normal force is smaller when pulling, the maximum static friction force is also smaller. Therefore, a smaller applied force $P$ is required to overcome friction when pulling.
A 20 kg crate is pulled across a rough horizontal floor by a rope that exerts a force of 100 N at an angle of 37° above the horizontal. If the coefficient of kinetic friction is 0.20, what is the magnitude of the friction force on the crate?
Answer: 27.2 N The friction force is $F_k = \mu_k F_N$ . First, we must find the normal force $F_N$ . The vertical forces must sum to zero. Let the applied force be $T$ . The upward forces are the normal force and the vertical component of the tension ( $T \sin\theta$ ). The downward force is gravity ( $mg$ ). So, $F_N + T \sin\theta - mg = 0$ , which means $F_N = mg - T \sin\theta$ . Using $g \approx 9.8 \, \text{m/s}^2$ , $\sin 37° \approx 0.6$ : $F_N = (20)(9.8) - (100)(0.6) = 196 - 60 = 136 \, \text{N}$ . Now, calculate the friction force: $F_k = (0.20)(136 \, \text{N}) = 27.2 \, \text{N}$ .
A box is sliding to the right on a horizontal floor with an initial velocity $v_0$ . The coefficient of kinetic friction between the box and the floor is $\mu_k$ . Which of the following statements correctly describes the kinetic friction force?
Answer: The friction force is directed to the left and has a magnitude of $\mu_k mg$ . The kinetic friction force always opposes the direction of relative motion. Since the box is sliding to the right, the friction force is directed to the left. The magnitude of the kinetic friction force is given by $F_k = \mu_k F_N$ . For an object on a horizontal surface, the normal force $F_N$ equals the weight $mg$ . Thus, the magnitude of the friction force is $\mu_k mg$ and is constant as long as the object is in motion and the coefficient of kinetic friction is constant.
A time-varying horizontal force $F(t) = ct$ is applied to a block of mass $m$ resting on a horizontal surface with a coefficient of static friction $\mu_s$ . Here, $c$ is a positive constant. At what time $t$ will the block begin to move?
Answer: $t = \mu_s mg / c$ The block will begin to move when the applied force $F(t)$ becomes equal to the maximum static friction force, $F_{s,max} = \mu_s F_N$ . For a block on a horizontal surface, $F_N = mg$ . So, the block starts to move when $F(t) = \mu_s mg$ . Since $F(t) = ct$ , we set $ct = \mu_s mg$ . Solving for $t$ gives $t = \mu_s mg / c$ .
A block slides down an inclined plane at constant velocity when the angle is $\theta_1 = 30°$ . The same block is then placed on an inclined plane at angle $\theta_2 = 45°$ . What is the acceleration of the block down the $45°$ incline?
Answer: $g(\sin 45° - \cos 45° \tan 30°)$ At constant velocity on the 30° incline: $mg\sin 30° = \mu_k mg\cos 30°$ , so $\mu_k = \tan 30°$ . On the 45° incline: $ma = mg\sin 45° - \mu_k mg\cos 45° = mg(\sin 45° - \cos 45° \tan 30°)$ . Therefore $a = g(\sin 45° - \cos 45° \tan 30°)$ . Option B incorrectly uses $1/\tan 30° = \cot 30°$ . Options C and D don't account for the friction coefficient properly.

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