Friction Practice Problems with Answers (AP Physics 1)

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Kinetic friction from a known normal force

   A crate slides across a level warehouse floor. The floor pushes up on the crate with a normal force of 240 N, and the coefficient of kinetic friction between the crate and the floor is $\mu_k = 0.28$. Find the magnitude of the kinetic friction force on the crate.

   Show the worked solution

   1. The crate is already sliding, so friction is kinetic. Kinetic friction is an equality rather than an inequality: it takes one value, fixed by the two surfaces and the normal force, no matter what else is pushing.

      $$\left|\vec{F}_{f,k}\right| = \left|\mu_k \vec{F}_N\right|$$

   2. Both inputs are given, so substitute straight in.

      $$F_{f,k} = (0.28)(240\ \text{N}) = 67.2\ \text{N}$$

   3. The coefficient carries two significant figures, so the answer stops there: $67\ \text{N}$. Its direction is opposite the crate's motion along the floor, which is what makes it slow the crate down.

   $F_{f,k} = 67\ \text{N}$, directed opposite the crate's velocity. The crate's mass never entered the calculation, because the normal force was handed to you instead of being computed from a weight.

2. 2. Kinetic friction when you are given the mass instead

   A 15 kg toolbox slides across a level concrete floor. Nothing touches it vertically except gravity and the floor. The coefficient of kinetic friction between the toolbox and the floor is $\mu_k = 0.42$. Find the normal force on the toolbox and the kinetic friction force acting on it.

   Show the worked solution

   1. Take positive as up for the vertical axis. The toolbox slides along the floor without leaving it, so its vertical acceleration is zero and the vertical forces cancel. The only two are the floor's normal force up and the weight down.

      $$F_N = mg = (15\ \text{kg})(9.8\ \text{m/s}^2) = 147\ \text{N}$$

   2. Feed that normal force into the kinetic friction relationship.

      $$F_{f,k} = \mu_k F_N = (0.42)(147\ \text{N}) = 61.74\ \text{N}$$

   3. Round to the two significant figures in $\mu_k$: $62\ \text{N}$.

   $F_N = 147\ \text{N}$ and $F_{f,k} = 62\ \text{N}$. Writing $F_N = mg$ was legitimate here only because the floor is level and nothing else acts on the toolbox vertically. Problem 5 breaks the second of those conditions and problem 6 breaks the first.

3. 3. Working backwards to the coefficient

   A force sensor drags a 3.4 kg wooden block horizontally across a level lab bench at a steady 0.20 m/s. The sensor reads a constant 8.6 N throughout. Find the coefficient of kinetic friction between the block and the bench.

   Show the worked solution

   1. Steady speed means zero acceleration, so the horizontal forces balance. Take positive as the direction of travel. The 8.6 N pull and kinetic friction are the only horizontal forces, so friction must match the pull in size.

      $$F_{f,k} = 8.6\ \text{N}$$

   2. The bench is level and the pull is horizontal, so the normal force is the whole weight.

      $$F_N = mg = (3.4\ \text{kg})(9.8\ \text{m/s}^2) = 33.32\ \text{N}$$

   3. Rearrange the kinetic friction relationship to isolate the coefficient.

      $$\mu_k = \frac{F_{f,k}}{F_N} = \frac{8.6\ \text{N}}{33.32\ \text{N}} = 0.2581$$

   4. Two significant figures: $\mu_k = 0.26$. It is a ratio of two forces, so the newtons cancel and the result has no units at all.

   $\mu_k = 0.26$, with no units. The 0.20 m/s was given only to tell you the block was moving at constant velocity. The speed itself never appears in the arithmetic, and dragging the block twice as fast would give the same coefficient.

4. 4. Does it move at all? Static versus kinetic

   A 26 kg crate sits at rest on a level floor. The coefficients between crate and floor are $\mu_s = 0.62$ and $\mu_k = 0.48$. (a) A worker pushes horizontally with 140 N. Does the crate start to move, and what friction force acts on it? (b) What is the smallest horizontal push that would get it moving? (c) With the crate now sliding, the worker settles to a steady 200 N horizontal push. Find the crate's acceleration.

   Show the worked solution

   1. Take positive as the direction of the push. The floor is level and the push is horizontal in every part, so the normal force is the weight throughout.

      $$F_N = mg = (26\ \text{kg})(9.8\ \text{m/s}^2) = 254.8\ \text{N}$$

   2. Part (a). Compare the push against the ceiling on static friction. That ceiling is the one place the static coefficient appears with an equals sign.

      $$F_{f,s,\text{max}} = \mu_s F_N = (0.62)(254.8\ \text{N}) = 158.0\ \text{N}$$

   3. The 140 N push is below the 158 N ceiling, so the crate stays at rest. Static friction is therefore not $\mu_s F_N$ here. It adopts whatever value keeps the crate still, and since the push is the only other horizontal force, that value is 140 N backwards. Quoting 158 N is the standard way to lose this part.

   4. Part (b). The crate breaks loose the moment the push exceeds what static friction can supply, so the smallest push that starts it moving is anything just above 158 N.

   5. Part (c). The crate is sliding now, so switch coefficients. Kinetic friction is fixed and no longer adapts.

      $$F_{f,k} = \mu_k F_N = (0.48)(254.8\ \text{N}) = 122.3\ \text{N}$$

   6. Newton's second law along the floor, with the push forwards and friction backwards.

      $$a = \frac{F_{\text{push}} - F_{f,k}}{m} = \frac{200\ \text{N} - 122.3\ \text{N}}{26\ \text{kg}} = \frac{77.70\ \text{N}}{26\ \text{kg}} = 2.988\ \text{m/s}^2$$

   7. Two significant figures: $3.0\ \text{m/s}^2$ in the direction of the push.

   (a) The crate does not move, and static friction is $140\ \text{N}$ backwards, not 158 N. (b) Any push above $158\ \text{N}$. (c) $a = 3.0\ \text{m/s}^2$ forwards. Friction on this crate drops from 158 N at the instant it breaks loose to 122 N once it is sliding, which is why a heavy crate lurches forwards as it starts.

5. 5. An angled pull, where the normal force is not mg

   A 9.0 kg box is dragged along a level floor by a rope held at $28^\circ$ above the horizontal. The tension in the rope is 45 N and the coefficient of kinetic friction between box and floor is $\mu_k = 0.22$. Find the normal force on the box, the kinetic friction force, and the box's acceleration.

   Show the worked solution

   1. Take positive $x$ as the direction of travel and positive $y$ as up. Split the rope tension into components.

      $$T_x = T\cos 28^\circ = (45\ \text{N})(0.8829) = 39.73\ \text{N}$$
      $$T_y = T\sin 28^\circ = (45\ \text{N})(0.4695) = 21.13\ \text{N}$$

   2. The box stays on the floor, so the vertical forces still cancel. There are now three of them rather than two: the normal force up, the weight down, and the rope's upward component.

      $$F_N + T_y - mg = 0$$

   3. The weight is $mg = (9.0\ \text{kg})(9.8\ \text{m/s}^2) = 88.2\ \text{N}$, so solve for the normal force.

      $$F_N = mg - T_y = 88.2\ \text{N} - 21.13\ \text{N} = 67.07\ \text{N}$$

      The rope is carrying part of the box's weight, so the floor has less left to carry.

   4. Friction responds to that smaller normal force, not to the weight.

      $$F_{f,k} = \mu_k F_N = (0.22)(67.07\ \text{N}) = 14.76\ \text{N}$$

   5. Newton's second law along the floor, using only the horizontal component of the tension.

      $$a = \frac{T_x - F_{f,k}}{m} = \frac{39.73\ \text{N} - 14.76\ \text{N}}{9.0\ \text{kg}} = \frac{24.97\ \text{N}}{9.0\ \text{kg}} = 2.8\ \text{m/s}^2$$

      Two significant figures is as far as the inputs support. Keeping full precision all the way through instead of the rounded values above gives $2.775\ \text{m/s}^2$, which rounds to the same $2.8\ \text{m/s}^2$.

   6. Compare that with the answer you would have got by writing $F_N = mg$ out of habit: friction would have come out as $(0.22)(88.2\ \text{N}) = 19.40\ \text{N}$, the net force as $20.33\ \text{N}$, and the acceleration as $2.3\ \text{m/s}^2$, about 19 percent low. Everything downstream of the normal force inherits the error.

   $F_N = 67\ \text{N}$, $F_{f,k} = 15\ \text{N}$, and $a = 2.8\ \text{m/s}^2$ in the direction of the pull. The normal force sits 21 N below the box's 88 N weight, and every friction number downstream inherits that. Angle the rope more steeply and the floor supports less still; angle the pull downwards instead and you would add the vertical component rather than subtract it.

6. 6. A block that stays put on a ramp

   A 7.5 kg box rests on a plank tilted at $21^\circ$ above the horizontal. The coefficient of static friction between box and plank is $\mu_s = 0.58$. Does the box slide? What friction force actually acts on it, and what is the largest friction force this surface could supply at this angle?

   Show the worked solution

   1. Use axes along and perpendicular to the plank. Positive is down the slope, and perpendicular positive points away from the surface. Resolve the weight, $mg = (7.5\ \text{kg})(9.8\ \text{m/s}^2) = 73.5\ \text{N}$, into those two directions.

      $$mg\sin\theta = (73.5\ \text{N})(\sin 21^\circ) = 26.34\ \text{N} \quad \text{down the slope}$$
      $$mg\cos\theta = (73.5\ \text{N})(\cos 21^\circ) = 68.62\ \text{N} \quad \text{into the surface}$$

   2. Perpendicular to the plank there is no acceleration, so the normal force matches the perpendicular part of the weight: $F_N = 68.62\ \text{N}$. It is not 73.5 N. On a ramp the surface only has to support the perpendicular component.

   3. Now find the ceiling on static friction at this angle.

      $$F_{f,s,\text{max}} = \mu_s F_N = (0.58)(68.62\ \text{N}) = 39.80\ \text{N}$$

   4. Compare the 26.34 N pulling the box down the slope with the 39.80 N the surface can supply. The surface can supply more than enough, so the box does not slide.

   5. Because it is not sliding, static friction is not 39.80 N. It takes whatever value keeps the box in equilibrium along the slope, and the only other along-slope force is the 26.34 N component of the weight, so

      $$F_{f,s} = 26.34\ \text{N} \quad \text{up the slope}$$

   6. A shortcut worth carrying into the next problem: divide the along-slope requirement by the perpendicular one and both $m$ and $g$ cancel, leaving $\tan\theta$ against $\mu_s$. Here $\tan 21^\circ = 0.38$, comfortably below 0.58, so the box holds whatever it weighs.

   The box stays put. Static friction is $26\ \text{N}$ acting up the slope, well under the $40\ \text{N}$ maximum this surface could have supplied. About 13 N of capacity goes unused, which is why the box would still hold if you loaded it or tilted the plank somewhat further.

7. 7. The critical angle, where the surface lets go

   A steel washer rests on a flat steel plate. The coefficient of static friction between them is $\mu_s = 0.74$. One edge of the plate is raised slowly from horizontal. At what angle does the washer begin to slide, and would stacking a second identical washer on top change that angle?

   Show the worked solution

   1. Positive is down the slope. At the instant sliding begins the washer is still in equilibrium, on the verge of moving, and static friction has just reached its ceiling for the first time. Set the along-slope pull equal to that ceiling.

      $$mg\sin\theta_c = \mu_s F_N = \mu_s\, mg\cos\theta_c$$

   2. Both $m$ and $g$ appear on both sides, so they cancel and the washer's weight drops out of the problem entirely.

      $$\frac{\sin\theta_c}{\cos\theta_c} = \tan\theta_c = \mu_s$$

   3. Solve for the angle.

      $$\theta_c = \arctan(0.74) = 36.50^\circ$$

   4. The coefficient carries two significant figures, so quoting about $37^\circ$ is the honest precision.

   5. Stacking a second washer doubles the weight, which doubles $mg\sin\theta$ and doubles $F_N$ at the same time. Both sides of the equilibrium equation double, so they still balance at the same angle. The critical angle is unchanged.

   The washer begins to slide at $\theta_c = 36.5^\circ$, about $37^\circ$ to two significant figures. A second stacked washer changes nothing, because $\tan\theta_c = \mu_s$ contains no mass. That independence is exactly why tilting a surface until an object slips is a standard way to measure $\mu_s$ with no scale and no force sensor.

8. 8. A block that does slide, and how fast it gets going

   An 11 kg block is released from rest on a ramp inclined at $42^\circ$ above the horizontal and slides down it. The coefficient of kinetic friction between block and ramp is $\mu_k = 0.31$. Find the friction force on the block, its acceleration, and its speed after it has slid 2.4 m along the ramp.

   Show the worked solution

   1. Positive is down the slope, which is the way the block is heading. Resolve the weight, $mg = (11\ \text{kg})(9.8\ \text{m/s}^2) = 107.8\ \text{N}$.

      $$mg\sin 42^\circ = (107.8\ \text{N})(0.6691) = 72.13\ \text{N}$$
      $$mg\cos 42^\circ = (107.8\ \text{N})(0.7431) = 80.11\ \text{N}$$

   2. There is no acceleration perpendicular to the ramp, so $F_N = 80.11\ \text{N}$. The block is sliding, so friction is kinetic and fixed at

      $$F_{f,k} = \mu_k F_N = (0.31)(80.11\ \text{N}) = 24.83\ \text{N}$$

      It acts up the slope, which is the negative direction here, because it opposes the block's motion.

   3. Newton's second law along the slope, down-slope positive.

      $$a = \frac{mg\sin\theta - F_{f,k}}{m} = \frac{72.13\ \text{N} - 24.83\ \text{N}}{11\ \text{kg}} = \frac{47.30\ \text{N}}{11\ \text{kg}} = 4.300\ \text{m/s}^2$$

   4. Doing the same algebra symbolically shows the mass cancelling, which is where the standard form comes from.

      $$a = g\left(\sin\theta - \mu_k\cos\theta\right) = 9.8\left(0.66913 - (0.31)(0.74314)\right)$$
      $$a = 9.8\left(0.43876\right) = 4.300\ \text{m/s}^2$$

   5. From here it is constant-acceleration kinematics: starting from rest, 2.4 m down the slope. Use the equation with no time in it, since no time was given.

      $$v^2 = v_0^2 + 2a\Delta x = 0 + 2(4.300\ \text{m/s}^2)(2.4\ \text{m}) = 20.64\ \text{m}^2/\text{s}^2$$
      $$v = \sqrt{20.64\ \text{m}^2/\text{s}^2} = 4.543\ \text{m/s}$$

   6. The distance and the coefficient both carry two significant figures, so $v = 4.5\ \text{m/s}$.

   $F_{f,k} = 25\ \text{N}$ up the slope, $a = 4.3\ \text{m/s}^2$ down the slope, and $v = 4.5\ \text{m/s}$ after 2.4 m. The acceleration does not depend on the 11 kg: any block with $\mu_k = 0.31$ on a $42^\circ$ ramp accelerates at $4.3\ \text{m/s}^2$. The mass was needed only for the friction force in newtons.

9. 9. Skidding to a stop: friction plus kinematics

   A car travelling at 22 m/s along a level road brakes so hard that the wheels lock and the tires skid. The coefficient of kinetic friction between the locked tires and the dry road is $\mu_k = 0.72$. How far does the car skid before it stops, and would a heavier car with identical tires skid further?

   Show the worked solution

   1. Take positive as the direction the car is travelling. The road is level and nothing else pushes the car horizontally, so friction is the only horizontal force, and it points backwards.

      $$F_N = mg, \qquad F_{f,k} = \mu_k mg$$

   2. Newton's second law along the road. The mass appears in the friction force and again in the inertia, so it cancels.

      $$a = \frac{-\mu_k mg}{m} = -\mu_k g = -(0.72)(9.8\ \text{m/s}^2) = -7.056\ \text{m/s}^2$$

   3. No time was given and none was asked for, so use the kinematic equation without $t$ in it.

      $$v^2 = v_0^2 + 2a\Delta x$$

   4. Set the final speed to zero and solve for the distance.

      $$0 = (22\ \text{m/s})^2 + 2(-7.056\ \text{m/s}^2)\Delta x$$
      $$\Delta x = \frac{484\ \text{m}^2/\text{s}^2}{14.11\ \text{m/s}^2} = 34.30\ \text{m}$$

   5. Two significant figures: the car skids about $34\ \text{m}$.

   The car skids $34\ \text{m}$. A heavier car with the same tires skids exactly as far, because the mass cancels: $\Delta x = v_0^2 / (2\mu_k g)$ contains no $m$. Speed is the variable that matters, and it is squared, so a car at 44 m/s would need four times the distance.

10. 10. Down a ramp, then along the floor

    A 3.8 kg block is released from rest at the top of a ramp inclined at $25^\circ$. On the ramp, $\mu_s = 0.35$ and $\mu_k = 0.28$. The block slides 2.5 m down the ramp and then continues onto a level floor, where the coefficient of kinetic friction is $\mu_k = 0.34$. Treat the join between ramp and floor as smooth, so the block keeps its speed as it arrives. How far along the floor does the block slide before it stops?

    Show the worked solution

    1. First check that it moves at all. On a ramp the block is on the verge of slipping when $\tan\theta = \mu_s$, so compare the two numbers: $\tan 25^\circ = 0.47$ against $\mu_s = 0.35$. The slope wins, so the block slides as soon as it is released, and every friction force from here on is kinetic.

    2. Ramp phase. Positive is the direction of travel, which on the ramp means down the slope. The block is sliding, so use the result derived in problem 8.

       $$a_1 = g\left(\sin\theta - \mu_k\cos\theta\right) = 9.8\left(\sin 25^\circ - (0.28)\cos 25^\circ\right)$$
       $$a_1 = 9.8\left(0.42262 - 0.25377\right) = 9.8(0.16885) = 1.655\ \text{m/s}^2$$

    3. Find the speed at the bottom, from rest over 2.5 m. Keep the answer as $v^2$ rather than $v$: the next phase needs the square anyway, so carrying it avoids a rounding step.

       $$v^2 = 2a_1 d_1 = 2(1.655\ \text{m/s}^2)(2.5\ \text{m}) = 8.275\ \text{m}^2/\text{s}^2$$

       That corresponds to $v = 2.88\ \text{m/s}$ arriving at the floor.

    4. Floor phase. Positive is still the direction of travel, now horizontal. The floor is level and nothing pulls on the block, so the normal force returns to the full weight, and the coefficient changes to the floor's 0.34. As in problem 9 the mass cancels.

       $$a_2 = -\mu_k g = -(0.34)(9.8\ \text{m/s}^2) = -3.332\ \text{m/s}^2$$

    5. Bring the block to rest with the no-time kinematic equation again.

       $$0 = v^2 + 2a_2 d_2 \quad \Rightarrow \quad d_2 = \frac{8.275\ \text{m}^2/\text{s}^2}{2(3.332\ \text{m/s}^2)} = 1.242\ \text{m}$$

    6. Two significant figures: about $1.2\ \text{m}$.

    The block slides about $1.2\ \text{m}$ along the floor. The 3.8 kg cancels at every stage and never affects the answer. Notice how differently the three coefficients are used: $\mu_s = 0.35$ only decides whether the block moves, while $\mu_k = 0.28$ and $\mu_k = 0.34$ decide how fast it gets going and how soon it stops.

11. 11. Explain why: the contact area is missing from the model

    A rectangular wooden block has one face measuring 20 cm by 10 cm and another measuring 20 cm by 5 cm. A student slides it across a bench at constant speed, first resting on the large face and then on the small face, and measures the same kinetic friction force both times even though the second footprint is half the area. Explain why the model predicts this, and state what the coefficient of kinetic friction does depend on.

    Show the worked solution

    1. Start by writing down what the model actually contains. Kinetic friction is $F_{f,k} = \mu_k F_N$. There are exactly two inputs, a coefficient and a normal force, and no area term anywhere. A model with no area in it cannot return a different answer for a different footprint.

    2. Check that neither input changed between the two runs. The bench is level and it is the same block, so $F_N = mg$ is identical both times. The two materials in contact are still wood and bench, so $\mu_k$ is identical too. Same inputs, same output.

    3. The CED states this outright. Essential knowledge 2.7.A.1.ii says that the force of friction between two surfaces does not depend on the size of the surface area of contact.

    4. The physical reason, which AP does not ask you to reproduce but which makes the result stop feeling arbitrary: surfaces touch only at microscopic high points, so the real area in contact is a small fraction of the footprint you can see. Halving the footprint doubles the pressure on those high points and flattens them further, and the real contact area ends up roughly where it started.

    5. Finally, what the coefficient does depend on. Essential knowledge 2.7.A.2.i says the coefficient of kinetic friction depends on the material properties of the surfaces in contact. Both surfaces, not one: there is no coefficient of friction for wood on its own, only for wood against a particular other material.

    Because $F_{f,k} = \mu_k F_N$ contains no area term, and neither of its two inputs changed when the block was turned over. The CED says so directly in essential knowledge 2.7.A.1.ii, that friction between two surfaces does not depend on the size of the contact area. The coefficient depends instead on the material properties of both surfaces in contact, per essential knowledge 2.7.A.2.i.