Newton's Laws — Physics Practice Worksheet

Grades 9–11 Medium difficulty 10 problems

Newton's First Law (inertia): an object at rest stays at rest and an object in motion stays in motion at constant velocity unless acted on by a net external force.

Newton's Second Law: Fnet=maF_{\mathrm{net}} = ma. Net force equals mass times acceleration. Force is measured in Newtons (N=kg⋅m/s2\mathrm{N} = \mathrm{kg \cdot m/s^2}). Direction of acceleration matches net force.

Newton's Third Law: for every action there is an equal and opposite reaction. Action-reaction pairs act on different objects and never cancel on the same object.

Skills practiced

  • Applying F=maF = ma
  • Identifying action-reaction pairs
  • Drawing and interpreting free-body diagrams
  • Distinguishing weight from mass
Name: Date: Period:

Practice Worksheet: Newton's Laws

Instructions: Solve each problem carefully. Show all work clearly. Write your final answer in the space provided or on a separate sheet as directed.

  1. 1.A 5.0 kg box is pushed across a frictionless floor with a constant force of 20 N. What is the acceleration of the box?
  2. 2.Which of the following best describes Newton's First Law of Motion?
    1. A. (a) Force equals mass times acceleration.
    2. B. (b) For every action, there is an equal and opposite reaction.
    3. C. (c) An object at rest stays at rest, and an object in motion stays in motion with the same speed and direction unless acted upon by an unbalanced force.
    4. D. (d) The acceleration of an object is directly proportional to the net force and inversely proportional to its mass.
  3. 3.A 12 kg object experiences a net force of 48 N. Calculate the acceleration of the object.
  4. 4.Label the forces acting on a book resting on a table. Draw and label arrows for the gravitational force (weight) and the normal force. Assume the book is stationary.
  5. 5.Two students push on a 150 kg cart. One pushes with 75 N to the right, and the other pushes with 45 N to the left. What is the net force on the cart? What is its acceleration?
  6. 6.According to Newton's Third Law, if you push against a wall with a force of 100 N, what force does the wall exert on you?
  7. 7.A 0.50 kg ball is thrown with an acceleration of 15 m/s2^2. What is the net force acting on the ball?
  8. 8.A car is moving at a constant velocity of 20 m/s on a straight, level road. Which of the following is true about the forces acting on the car?
    1. A. (a) The net force is zero.
    2. B. (b) The net force is in the direction of motion.
    3. C. (c) The net force is opposite the direction of motion.
    4. D. (d) The net force is increasing.
  9. 9.A 60 kg person is standing on a scale in an elevator. The elevator is accelerating upward at 2.0 m/s2^2. What is the reading on the scale? (Use g=9.8g = 9.8 m/s2^2.)
  10. 10.A 10 kg crate is pulled across a rough floor with a force of 50 N at an angle of 30∘^\circ above the horizontal. The frictional force opposing motion is 15 N. Draw a free-body diagram showing all forces acting on the crate. Then calculate the net horizontal force on the crate.

Answer Key

  1. 1.
    F=ma⇒a=Fm=20 N5.0 kg=4.0 m/s2F = ma \Rightarrow a = \frac{F}{m} = \frac{20\ \text{N}}{5.0\ \text{kg}} = 4.0\ \text{m/s}^2
    Final answer: 4.0 m/s24.0\ \text{m/s}^2
  2. 2.
    Final answer: (c) An object at rest stays at rest, and an object in motion stays in motion with the same speed and direction unless acted upon by an unbalanced force.
  3. 3.
    F=ma⇒a=Fm=48 N12 kg=4.0 m/s2F = ma \Rightarrow a = \frac{F}{m} = \frac{48\ \text{N}}{12\ \text{kg}} = 4.0\ \text{m/s}^2
    Final answer: 4.0 m/s24.0\ \text{m/s}^2
  4. 4.
    Final answer: Diagram should show: an upward arrow labeled "Normal Force" from the table on the book, and a downward arrow labeled "Weight" (or FgF_g) from the center of the book. Both arrows are equal in length.
  5. 5.
    Net force: 75 N−45 N=30 N75\ \text{N} - 45\ \text{N} = 30\ \text{N} to the right. Acceleration: a=Fnetm=30 N150 kg=0.20 m/s2a = \frac{F_{\text{net}}}{m} = \frac{30\ \text{N}}{150\ \text{kg}} = 0.20\ \text{m/s}^2 to the right.
    Final answer: Net force: 30 N30\ \text{N} to the right. Acceleration: 0.20 m/s20.20\ \text{m/s}^2 to the right.
  6. 6.
    Final answer: The wall exerts a force of 100 N back on the person, in the opposite direction.
  7. 7.
    F=ma=(0.50 kg)(15 m/s2)=7.5 NF = ma = (0.50\ \text{kg})(15\ \text{m/s}^2) = 7.5\ \text{N}
    Final answer: 7.5 N7.5\ \text{N}
  8. 8.
    Final answer: (a) The net force is zero.
  9. 9.
    The scale reads the normal force. Fnet=ma=(60 kg)(2.0 m/s2)=120 NF_{\text{net}} = ma = (60\ \text{kg})(2.0\ \text{m/s}^2) = 120\ \text{N} upward. Fnet=FN−mg⇒FN=mg+Fnet=(60 kg)(9.8 m/s2)+120 N=588 N+120 N=708 NF_{\text{net}} = F_N - mg \Rightarrow F_N = mg + F_{\text{net}} = (60\ \text{kg})(9.8\ \text{m/s}^2) + 120\ \text{N} = 588\ \text{N} + 120\ \text{N} = 708\ \text{N}
    Final answer: 708 N708\ \text{N}
  10. 10.
    Free-body diagram: weight down, normal force up, applied force at 30∘^\circ above horizontal (rightward), friction leftward. Net horizontal force: Fapplied,x=50 Ncos⁡30∘≈43.3 NF_{\text{applied},x} = 50\ \text{N} \cos 30^\circ \approx 43.3\ \text{N}. Fnet,x=43.3 N−15 N=28.3 NF_{\text{net},x} = 43.3\ \text{N} - 15\ \text{N} = 28.3\ \text{N} to the right.
    Final answer: 28.3 N28.3\ \text{N} to the right.
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Common mistakes to avoid

  • Using total force instead of net force in F=maF = ma
  • Thinking action-reaction forces cancel on one object
  • Confusing mass (kg) with weight (N)
  • Forgetting friction opposes motion

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