What is Forces & dynamics on the Physics Regents?

Forces & dynamics is a Mechanics topic on the Physics Regents. RegentsLab has 69 practice questions on it, each with a full worked solution.

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Forces & dynamics: Physics Regents Practice

Practice Forces & dynamics for the Physics Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 17 · 2 pts

A net force of $10.\ \text{N}$ acts on a $2.0\ \text{kg}$ cart resting on a frictionless horizontal surface. What is the magnitude of the cart's acceleration?

(A) $12\ \text{m/s}^2$
(B) $20.\ \text{m/s}^2$
(C) $5.0\ \text{m/s}^2$ ✓
(D) $0.20\ \text{m/s}^2$

Worked solution

$F_{net} = ma \;\Rightarrow\; a = \frac{F_{net}}{m}$ Newton's second law from the Reference Tables.

$a = \frac{10.\ \text{N}}{2.0\ \text{kg}}$

$a = 5.0\ \text{m/s}^2$

Answer: $5.0\ \text{m/s}^2$

Newton's second law states acceleration equals net force divided by mass, so a larger mass under the same force accelerates less.

Question 18 · 2 pts

What is the weight (force of gravity) acting on a $5.0\ \text{kg}$ mass near the surface of Earth?

(A) $5.0\ \text{N}$
(B) $0.51\ \text{N}$
(C) $9.8\ \text{N}$
(D) $49\ \text{N}$ ✓

Worked solution

$F_g = mg$ Weight equation; g = 9.81 m/s^2 from the Reference Tables.

$F_g = (5.0\ \text{kg})(9.81\ \text{m/s}^2)$

$F_g = 49\ \text{N}$

Answer: $49\ \text{N}$

Weight is the gravitational force on a mass, equal to the product of mass and the acceleration due to gravity $g$.

Question 25 · 2 pts

On Earth, what is the weight of an object with a mass of $10\ \mathrm{kg}$? (Use $g = 9.81\ \mathrm{m/s^2}$ from the reference tables.)

(A) $19.81\ \mathrm{N}$
(B) $1.02\ \mathrm{N}$
(C) $10\ \mathrm{N}$
(D) $98.1\ \mathrm{N}$ ✓

Worked solution

$F_g = mg$ Weight equals mass times gravitational field strength.

$F_g = (10)(9.81) = 98.1\ \mathrm{N}$

Answer: $98.1\ \mathrm{N}$

Weight is the gravitational force on a mass, $F_g = mg$, using $g = 9.81\ \mathrm{m/s^2}$ near Earth's surface.

Question 26 · 2 pts

A spring with a spring constant of $150\ \mathrm{N/m}$ is stretched a distance of $0.20\ \mathrm{m}$. What is the force exerted by the spring? (Use $F_s = kx$.)

(A) $750\ \mathrm{N}$
(B) $150.2\ \mathrm{N}$
(C) $0.0013\ \mathrm{N}$
(D) $30\ \mathrm{N}$ ✓

Worked solution

$F_s = kx$ Hooke's law for an ideal spring.

$F_s = (150)(0.20) = 30\ \mathrm{N}$

Answer: $30\ \mathrm{N}$

An ideal spring's force is proportional to its stretch: $F_s = kx$, where $k$ is the spring constant.

Question 3 · 2 pts

What net force is required to give a $5.0\ \text{kg}$ object an acceleration of $2.0\ \text{m/s}^2$?

(A) $7.0\ \text{N}$
(B) $3.0\ \text{N}$
(C) $10\ \text{N}$ ✓
(D) $2.5\ \text{N}$

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\ \text{kg})(2.0\ \text{m/s}^2)$

$F_{net} = 10\ \text{N}$

Answer: $10\ \text{N}$

Newton's second law states net force equals mass times acceleration; multiplying the given mass and acceleration gives 10 N.

Question 15 · 2 pts

A net force acts on a $5.0\text{-kg}$ object, giving it an acceleration of $3.0\text{ m/s}^2$. What is the magnitude of the net force acting on the object?

(A) $1.7\text{ N}$
(B) $8.0\text{ N}$
(C) $49\text{ N}$
(D) $15\text{ N}$ ✓

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\text{ kg})(3.0\text{ m/s}^2)$

$F_{net} = 15\text{ N}$

Answer: $15\text{ N}$

Newton's second law states that net force equals mass times acceleration. Multiplying the given mass and acceleration gives 15 N.

Question 27 · 2 pts

What net force is required to give a 5.0-kilogram object an acceleration of $3.0\ \text{m/s}^2$?

(A) 49 N
(B) 8.0 N
(C) 15 N ✓
(D) 1.7 N

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\ \text{kg})(3.0\ \text{m/s}^2)$

$F_{net} = 15\ \text{N}$

Answer: 15 N

Newton's second law states that net force equals mass times acceleration. Multiplying the given mass and acceleration gives 15 N; the 49 N distractor comes from mistakenly calculating the object's weight instead of the required net force.

Question 36 · 2 pts

A 2.0-kilogram book rests motionless on a horizontal table. What is the magnitude of the normal force exerted by the table on the book?

(A) 9.8 N
(B) 20. N ✓
(C) 0 N
(D) 2.0 N

Worked solution

$\Sigma F = 0$ Book is in equilibrium, so normal force balances weight

$F_N = mg$

$F_N = (2.0\ \text{kg})(9.81\ \text{m/s}^2)$

$F_N = 19.6\ \text{N} \approx 20.\ \text{N}$

Answer: 20. N

Since the book is not accelerating, the normal force from the table must balance the book's weight, mg. This gives about 20. N; the 2.0 N distractor mistakes the mass in kilograms for the force in newtons.

Question 41 · 2 pts

A 5.0 kg cart experiences a net force of 15 N. What is the magnitude of the cart's acceleration?

(A) 3.0 m/s^2 ✓
(B) 20 m/s^2
(C) 75 m/s^2
(D) 0.33 m/s^2

Worked solution

$F_{net} = m a \Rightarrow a = \dfrac{F_{net}}{m}$

$a = \dfrac{15\ \text{N}}{5.0\ \text{kg}} = 3.0\ \text{m/s}^2$

Answer: 3.0 m/s^2

Newton's second law relates net force, mass, and acceleration; dividing the net force by the mass gives 3.0 m/s^2. Inverting the ratio or adding the values instead of dividing produces the distractors.

Question 3 · 2 pts

What net force is required to give a 1500 kg car an acceleration of 2.0 m/s^2?

(A) 3.0 x 10^3 N ✓
(B) 6.0 x 10^3 N
(C) 1.5 x 10^3 N
(D) 7.5 x 10^2 N

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (1500\text{ kg})(2.0\text{ m/s}^2)$

$F_{net} = 3.0\times10^3\text{ N}$

Answer: 3.0 x 10^3 N

Newton's second law states that net force equals mass times acceleration. Multiplying the given mass and acceleration directly gives 3.0 x 10^3 N; the distractors come from dividing instead of multiplying, or from arithmetic slips.

Question 15 · 2 pts

A net force of $24\ \text{N}$ acts on a $6.0\ \text{kg}$ cart that starts at rest. What is the cart's speed after $3.0\ \text{s}$?

(A) 0.75 m/s
(B) 72 m/s
(C) 12 m/s ✓
(D) 4.0 m/s

Worked solution

$a = \dfrac{F_{net}}{m} = \dfrac{24\ \text{N}}{6.0\ \text{kg}} = 4.0\ \text{m/s}^2$

$v = v_0 + at = 0 + (4.0\ \text{m/s}^2)(3.0\ \text{s})$

$v = 12\ \text{m/s}$

Answer: 12 m/s

Newton's second law gives the acceleration, which is then used with the kinematics equation for constant acceleration to find final speed. Stopping after finding acceleration (4.0 m/s^2) is a common error since it has similar-looking units.

Question 39 · 2 pts

What net force is required to give a 5.0 kg mass an acceleration of 2.0 m/s^2?

(A) 2.5 N
(B) 10. N ✓
(C) 7.0 N
(D) 20. N

Worked solution

$F_{net} = ma$

$F_{net} = (5.0\text{ kg})(2.0\text{ m/s}^2)$

$F_{net} = 10.\text{ N}$

Answer: 10. N

Newton's second law states that net force equals mass times acceleration; multiplying 5.0 kg by 2.0 m/s^2 gives 10. N.

Question 3 · 2 pts

A net force is applied to a $5.0\ \text{kg}$ object, giving it an acceleration of $3.0\ \text{m/s}^2$. What is the magnitude of the net force?

(A) $45\ \text{N}$
(B) $8.0\ \text{N}$
(C) $1.7\ \text{N}$
(D) $15\ \text{N}$ ✓

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\ \text{kg})(3.0\ \text{m/s}^2)$

$F_{net} = 15\ \text{N}$

Answer: $15\ \text{N}$

Newton's second law states that net force equals mass times acceleration; multiplying the given mass and acceleration gives 15 N.

Question 18 · 2 pts

A $5.0\ \text{kg}$ block accelerates at $2.0\ \text{m/s}^2$ across a frictionless surface. What is the magnitude of the net force acting on the block?

(A) $10.\ \text{N}$ ✓
(B) $2.5\ \text{N}$
(C) $3.0\ \text{N}$
(D) $7.0\ \text{N}$

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\ \text{kg})(2.0\ \text{m/s}^2)$

$F_{net} = 10.\ \text{N}$

Answer: $10.\ \text{N}$

Newton's second law states net force equals mass times acceleration; multiplying the given mass and acceleration gives the net force directly.

Question 31 · 2 pts

A net force accelerates a 5.0 kg cart at $2.0\ \text{m/s}^2$ along a frictionless surface. What is the magnitude of the net force?

(A) 7.0 N
(B) 2.5 N
(C) 0.40 N
(D) 10. N ✓

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\ \text{kg})(2.0\ \text{m/s}^2)$

$F_{net} = 10.\ \text{N}$

Answer: 10. N

Newton's second law states net force equals mass times acceleration, so multiplying 5.0 kg by 2.0 m/s^2 gives 10. N. The distractor 2.5 N comes from dividing instead of multiplying, and 7.0 N comes from mistakenly adding mass and acceleration.

Question 43 · 2 pts

A net force acts on a 5.0 kg object, giving it an acceleration of $3.0\ \text{m/s}^2$. What is the magnitude of the net force?

(A) 8.0 N
(B) 45 N
(C) 15 N ✓
(D) 1.7 N

Worked solution

$F_{net} = ma$ Newton's second law

$F_{net} = (5.0\ \text{kg})(3.0\ \text{m/s}^2) = 15\ \text{N}$

Answer: 15 N

Newton's second law states that net force equals mass times acceleration. Multiplying 5.0 kg by 3.0 m/s^2 gives 15 N. Dividing instead of multiplying (1.7 N) or adding the values (8.0 N) are common arithmetic errors.

Question 2 · 2 pts

A 2.0 kg object experiences two horizontal forces: 10. N directed east and 6.0 N directed west. What is the magnitude of the object's acceleration?

(A) 3.0 m/s^2
(B) 2.0 m/s^2 ✓
(C) 8.0 m/s^2
(D) 0.5 m/s^2

Worked solution

$F_{net} = 10.\ N - 6.0\ N = 4.0\ N\ \text{(east)}$ subtract opposing forces to get net force

$a = F_{net}/m = 4.0/2.0 = 2.0\ m/s^2$

Answer: 2.0 m/s^2

The net force is the vector sum of the two opposing forces (4.0 N east); dividing by mass gives the acceleration via Newton's second law.

Question 16 · 2 pts

A net force is applied to a $5.0\ \text{kg}$ object, giving it an acceleration of $2.0\ \text{m/s}^2$. What is the magnitude of the net force?

(A) 2.5 N
(B) 49 N
(C) 7.0 N
(D) 10. N ✓

Worked solution

$F_{net} = ma = (5.0\ \text{kg})(2.0\ \text{m/s}^2) = 10.\ \text{N}$

Answer: 10. N

Newton's second law states that net force equals mass times acceleration. The 49 N distractor results from incorrectly multiplying mass by g instead of the given acceleration.

Question 42 · 2 pts

A 5.0-kilogram object accelerates at 3.0 meters per second squared. What is the magnitude of the net force acting on the object?

(A) 1.7 N
(B) 8.0 N
(C) 60 N
(D) 15 N ✓

Worked solution

$F_{net} = ma = (5.0\ \text{kg})(3.0\ \text{m/s}^2) = 15\ \text{N}$ Apply Newton's second law

Answer: 15 N

Newton's second law states that net force equals mass times acceleration, giving 15 N.

Question 61 · 4 pts · constructed response

A 5.0-kg block is pulled at constant velocity across a horizontal surface by a horizontal force of 15 N. Calculate the normal force on the block. Calculate the coefficient of kinetic friction between the block and the surface.

Model answer: The normal force is 49 N and the coefficient of kinetic friction is 0.31.

Worked solution

$F_N = mg = (5.0\ \text{kg})(9.81\ \text{m/s}^2)$ On a horizontal surface, normal force balances weight

$F_N \approx 49\ \text{N}$

$\text{Since velocity is constant, } F_{applied} = F_{friction} = 15\ \text{N}$ Constant velocity means net force is zero

$\mu_k = \dfrac{F_{friction}}{F_N} = \dfrac{15\ \text{N}}{49\ \text{N}} \approx 0.31$

Answer: The normal force is 49 N and the coefficient of kinetic friction is 0.31.

Because the block moves at constant velocity, the net force is zero, so the applied force equals the friction force, allowing the coefficient of kinetic friction to be calculated as the ratio of friction force to normal force.

Question 12 · 1 pt

A net force of $20\ \mathrm{N}$ acts on a $4\ \mathrm{kg}$ object. What is its acceleration?

(A) $5\ \mathrm{m/s^2}$ ✓
(B) $80\ \mathrm{m/s^2}$
(C) $0.2\ \mathrm{m/s^2}$
(D) $24\ \mathrm{m/s^2}$

Worked solution

$F = ma \Rightarrow a = \frac{F}{m}$ Newton's second law.

$= \frac{20}{4} = 5\ \mathrm{m/s^2}$

Answer: 5\ \mathrm{m/s^2}

Newton's second law: acceleration equals net force divided by mass.

Question 19 · 2 pts

A $10.\ \text{kg}$ box rests on a horizontal floor where the coefficient of friction is $0.30$. What is the magnitude of the maximum force of friction acting on the box?

(A) $98\ \text{N}$
(B) $3.0\ \text{N}$
(C) $29\ \text{N}$ ✓
(D) $327\ \text{N}$

Worked solution

$F_N = mg = (10.\ \text{kg})(9.81\ \text{m/s}^2) = 98.1\ \text{N}$ On a level surface the normal force equals the weight.

$F_f = \mu F_N$

$F_f = (0.30)(98.1\ \text{N}) = 29\ \text{N}$

Answer: $29\ \text{N}$

Friction force equals the coefficient of friction times the normal force, and on a level floor the normal force equals the object's weight.

Question 20 · 2 pts

Two forces act concurrently on a point object: $3.0\ \text{N}$ directed east and $4.0\ \text{N}$ directed north. What is the magnitude of the resultant (net) force?

(A) $25\ \text{N}$
(B) $1.0\ \text{N}$
(C) $5.0\ \text{N}$ ✓
(D) $7.0\ \text{N}$

Worked solution

$F_R = \sqrt{F_x^2 + F_y^2}$ Perpendicular vectors add by the Pythagorean theorem.

$F_R = \sqrt{(3.0\ \text{N})^2 + (4.0\ \text{N})^2} = \sqrt{9.0 + 16}$

$F_R = \sqrt{25} = 5.0\ \text{N}$

Answer: $5.0\ \text{N}$

Perpendicular force vectors combine by the Pythagorean theorem, not by simple addition, giving a resultant smaller than the scalar sum.

Question 23 · 2 pts

A car tows a trailer, and both speed up together along a straight road. While the car pulls forward on the trailer with a force of $1500\ \text{N}$, how does the force the trailer exerts on the car compare?

(A) It is zero because only the car has an engine to produce a force.
(B) It is greater than $1500\ \text{N}$ because the car is more massive than the trailer.
(C) It is $1500\ \text{N}$ directed backward, equal in magnitude and opposite in direction. ✓
(D) It is less than $1500\ \text{N}$ because the trailer is lighter and is being pulled.

Worked solution

$F_{A \text{ on } B} = -F_{B \text{ on } A}$ Newton's third law: forces come in equal, opposite action-reaction pairs.

$|F_{trailer \text{ on } car}| = |F_{car \text{ on } trailer}| = 1500\ \text{N}$ Equal magnitudes regardless of mass or motion.

$\text{Direction is opposite: the trailer pulls backward on the car.}$

Answer: $1500\ \text{N}$ directed backward (equal and opposite)

By Newton's third law, the action-reaction forces between two interacting objects are always equal in magnitude and opposite in direction, independent of their masses or motion.

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