Force and Translational Dynamics
Unit 2 of AP Physics 1, worth 18–23% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Systems and centre of mass, forces and free-body diagrams, Newton's three laws, gravitational force, friction, springs, circular motion.
How this unit is tested
Every problem in this unit starts the same way: define your system (one object, several objects, or a whole system whose center of mass you track), then draw a free-body diagram showing only the forces acting ON that system from the outside. Internal forces between parts of a system (like tension in a string connecting two blocks you've chosen to treat as one system) cancel out and never appear on the system's own diagram — this is the key idea behind 'systems and center of mass' problems, and it's what lets you skip solving for internal forces when you only need the overall acceleration.
Once the diagram is drawn, choose axes (usually along and perpendicular to the direction of acceleration, which is along an incline if there is one) and apply Newton's second law separately along each axis: $\sum F_x = ma_x$ and $\sum F_y = ma_y$. Perpendicular to the direction of motion, acceleration is usually zero, which is how you solve for the normal force. That normal force then feeds into friction and, on inclines, into the component of gravity you use.
Treat friction, springs, and circular motion as special cases of the same second-law approach, not as separate topics. Friction is a force whose maximum static value or fixed kinetic value you compute from $\mu N$ and then plug into $\sum F = ma$ like any other force. A spring's force is just $-kx$ plugged into the same equation. Circular motion is not a new force at all — 'centripetal force' is just the name for whatever net force (tension, friction, gravity, normal force, or a combination) happens to point toward the center and satisfies $F_{net} = mv^2/r$.
On the exam, expect multi-step problems that combine two or three of these ideas (an incline with friction, a two-block system with a hanging mass, a car rounding a banked curve). Always finish by checking that your answer's sign and direction make physical sense — does the acceleration point the way the net force points?
What you have to know
Newton's Second Law
The net force on an object equals its mass times its acceleration: $\sum F = ma$, applied independently along each perpendicular axis.
Newton's Third Law
If object A exerts a force on object B, object B exerts a force of equal magnitude and opposite direction on A. The two forces act on different objects and are of the same type.
Newton's Law of Universal Gravitation
Any two masses attract each other with force $F = \dfrac{Gm_1m_2}{r^2}$, where $r$ is the distance between their centers and $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$.
Static and Kinetic Friction
Static friction adjusts up to a maximum, $f_s \le \mu_s N$, to prevent relative sliding. Once sliding occurs, kinetic friction has a fixed magnitude $f_k = \mu_k N$, opposing relative motion, with $\mu_k \le \mu_s$ typically.
Hooke's Law
An ideal spring exerts a restoring force proportional to its displacement from equilibrium: $F_s = -kx$, where $k$ is the spring constant and the negative sign shows the force opposes the displacement.
Center of Mass Motion of a System
For any system of objects, $\sum F_{ext} = M_{total}\, a_{cm}$; internal forces between parts of the system do not affect the center of mass's acceleration.
14 practice questions
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A net force of 12 N acts on a 3 kg object initially at rest. What is its speed after 4 seconds?
- 8 m/s
- 12 m/s
- 16 m/s
- 48 m/s
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Answer. 16 m/s
By Newton's second law, $a = F/m = 12/3 = 4\ \text{m/s}^2$. Starting from rest, $v = at = 4(4) = 16\ \text{m/s}$. -
A 2 kg block and a 3 kg block are in contact on a frictionless surface. A 10 N horizontal force pushes on the 2 kg block, which pushes into the 3 kg block. What is the contact force between the blocks?
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Answer. 6 N
Treat both blocks as one system: $a = F/M = 10/5 = 2\ \text{m/s}^2$. Isolating the 3 kg block, the only horizontal force on it is the contact force from the 2 kg block, so $F_{contact} = m a = 3(2) = 6\ \text{N}$. -
A book rests on a table. Which of the following is the Newton's third law pair to the gravitational force of Earth on the book?
- The normal force of the table on the book
- The gravitational force of the book on Earth
- The normal force of the book on the table
- The weight of the table itself
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Answer. The gravitational force of the book on Earth
Third-law pairs act on different objects and are the same type of force. Earth pulls the book down with gravity, so the book must pull Earth up with an equal gravitational force; the normal force is a different type of force acting on a different pair of objects. -
A 5 kg block slides down a frictionless incline angled 30° above horizontal. What is its acceleration along the incline?
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Answer. about 4.9 m/s^2, directed down the incline
On a frictionless incline, only the component of gravity along the surface accelerates the block: $a = g\sin\theta = 9.8 \times 0.5 = 4.9\ \text{m/s}^2$. Mass does not affect this result. -
A 10 kg crate sits on a horizontal surface with $\mu_s = 0.4$ and $\mu_k = 0.3$. A horizontal 30 N force is applied. Does the crate move, and what is its acceleration?
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Answer. It does not move; acceleration is 0
Maximum static friction is $\mu_s m g = 0.4(10)(9.8) = 39.2\ \text{N}$, which exceeds the applied 30 N. Static friction simply matches the applied force to keep the crate at rest, so it does not slide. -
The same 10 kg crate ($\mu_k = 0.3$) now has a horizontal 45 N force applied and is sliding. What is its acceleration?
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Answer. about 1.56 m/s^2
Kinetic friction is $\mu_k m g = 0.3(10)(9.8) = 29.4\ \text{N}$, opposing motion. Net force is $45 - 29.4 = 15.6\ \text{N}$, so $a = 15.6/10 = 1.56\ \text{m/s}^2$. -
Two point masses, 2 kg and 3 kg, are separated by 0.5 m. What is the gravitational force between them?
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Answer. about 1.6 x 10^-9 N
Using $F = Gm_1m_2/r^2 = (6.67\times10^{-11})(2)(3)/(0.5)^2$, the numerator is $4.0\times10^{-10}$ and dividing by $0.25$ gives about $1.6\times10^{-9}\ \text{N}$. -
At Earth's surface, $g = 9.8\ \text{m/s}^2$. What is the gravitational field strength at a distance from Earth's center equal to twice Earth's radius?
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Answer. 2.45 m/s^2
Gravitational field strength follows $g \propto 1/r^2$. Doubling the distance from the center reduces $g$ by a factor of $2^2 = 4$, so $g' = 9.8/4 = 2.45\ \text{m/s}^2$. -
A spring with constant $k = 200\ \text{N/m}$ is stretched 0.15 m from equilibrium. What is the magnitude of the spring's restoring force?
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Answer. 30 N
By Hooke's Law, the magnitude is $F = kx = 200(0.15) = 30\ \text{N}$, directed opposite the stretch, back toward the equilibrium position. -
A 0.5 kg mass on a frictionless horizontal surface is attached to a spring with $k = 200\ \text{N/m}$ and displaced 0.1 m from equilibrium. What is the magnitude of its acceleration at that instant?
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Answer. 40 m/s^2
The spring force is $F = kx = 200(0.1) = 20\ \text{N}$. By Newton's second law, $a = F/m = 20/0.5 = 40\ \text{m/s}^2$, directed back toward equilibrium. -
A car rounds a flat, unbanked curve of radius 50 m at 20 m/s. What is the minimum coefficient of static friction needed between the tires and road to prevent sliding?
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Answer. about 0.82
Centripetal acceleration is $a_c = v^2/r = 400/50 = 8\ \text{m/s}^2$. Friction alone supplies this, so $\mu_s mg \ge ma_c$ gives $\mu_s \ge a_c/g = 8/9.8 \approx 0.82$. -
A 0.2 kg ball on a 0.5 m string moves in a vertical circle. What is the minimum speed at the top of the loop for the string tension to remain non-negative?
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Answer. about 2.21 m/s
At the minimum speed, tension is zero and gravity alone provides the centripetal force: $mg = mv^2/r$, so $v = \sqrt{gr} = \sqrt{9.8(0.5)} \approx 2.21\ \text{m/s}$. Mass cancels out. -
Two carts, 3 kg and 2 kg, are connected by a spring and sit on a frictionless track. An external 20 N horizontal force is applied only to the 3 kg cart. What is the acceleration of the system's center of mass?
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Answer. 4 m/s^2
The spring force between the carts is internal to the two-cart system and cancels when computing center of mass motion. Only the external 20 N matters: $a_{cm} = F_{ext}/M_{total} = 20/5 = 4\ \text{m/s}^2$. -
Which of the following statements about static and kinetic friction is correct?
- Kinetic friction is always greater than static friction
- The maximum static friction force is generally greater than or equal to the kinetic friction force for the same surfaces
- Friction force depends on the area of contact between surfaces
- Friction force is independent of the normal force
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Answer. The maximum static friction force is generally greater than or equal to the kinetic friction force for the same surfaces
Static friction can rise up to $\mu_s N$ to prevent sliding, and experimentally $\mu_s \ge \mu_k$ for most surface pairs, which is why more force is often needed to start motion than to sustain it. Friction is independent of contact area and depends on the normal force, not surface area.
What people get wrong
- Using $mg$ as the normal force on an incline instead of $mg\cos\theta$. Always find the normal force by balancing forces perpendicular to the incline's surface, not the horizontal.
- Applying $f_k = \mu_kN$ to an object that is not actually sliding. Check whether the applied force exceeds $\mu_sN$ first; if it doesn't, the object stays at rest and friction equals the applied force, not $\mu_kN$.
- Pairing the normal force and weight as a Newton's third law pair because they're equal and opposite. They act on the same object and are different types of force, so they cannot be a third-law pair; true pairs act on different objects.
- Treating 'centripetal force' as an extra force to add to the free-body diagram. It is not a new force — it's the label for the net force (or the net radial component of existing forces) that must point toward the center.
- Forgetting that tension and other contact forces between parts of a system cancel when you treat the whole system together, and re-deriving them unnecessarily instead of switching to a single-object diagram only when the internal force is actually asked for.
- Dropping the negative sign's meaning in Hooke's Law and getting the direction of the spring force wrong; the force always points back toward equilibrium, opposite the displacement.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.