The problems are on the Newton's laws worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Newton's laws link force and motion. The first law says zero net force means constant velocity, the second says ΣF = ma, and the third says forces come in equal and opposite pairs acting on different objects. These worked answers correct the 20 problems of the AP Physics 1 worksheet on Newton's laws, from forces on a book to truck against mosquito: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the chosen axis, the equation, the substitution and the result in bold with its unit. Check your signs first, then the numbers, because a wrong sign changes the answer by a large amount.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 2
Problem 1 Forces on a book
1. Firstly, weight (gravitational force of Earth on the book) downward and the normal force of the table on the book upward.2. Then the net force is zero, so the book stays at rest.
3. Finally, the force of the book on the table, directed downward.
Problem 2 Net force and acceleration
1. Firstly, a = 3000 ÷ 1200 = 2.5 m/s².2. v = at = 2.5 × 6.0 = 15 m/s.
Problem 3 Weight on two worlds
1. Firstly, W = 70 × 9.81 = 687 N.2. W = 70 × 1.62 = 113 N.
3. Finally, mass does not depend on location: 70 kg.
Problem 4 Forces in a line
1. Firstly, net force = 30 − 12 − 8 = 10 N to the right.2. a = 10 ÷ 5.0 = 2.0 m/s² to the right.
Problem 5 Bus brakes suddenly
1. Firstly, the passengers have no net force on them in the forward direction once the bus slows, so they tend to keep moving forward at their original velocity (first law).2. Then friction between the road and the brakes and tyres slows the bus. The passengers continue forward until a force from the floor, a rail or a seat acts on them: inertia, not a new forward force.
Problem 6 Perpendicular forces
1. Firstly, F = √(6.0² + 8.0²) = 10 N.2. a = 10 ÷ 2.0 = 5.0 m/s².
3. tan θ = 8.0 ÷ 6.0, so θ = 53° north of east.
Problem 7 Hanging mass
1. Firstly, T = mg = 4.0 × 9.81 = 39 N.2. T − mg = ma, T = 4.0 × (9.81 + 2.0) = 47 N.
3. T = 4.0 × (9.81 − 2.0) = 31 N.
Problem 8 Reading a bathroom scale
1. Firstly, W = 50 × 9.81 = 491 N.2. N − mg = ma, so a = (560 − 490.5) ÷ 50 = 1.4 m/s² upward.
3. Finally, the elevator is either speeding up while moving upward or slowing down while moving downward.
Problem 9 Third-law partners
1. Firstly, the partner is the force of Diego’s hand on the wall, in the opposite direction, of the same contact type.2. Then the partner is the force of the exhaust gas on the rocket, directed upward.
3. Finally, the partner is the gravitational force of the apple on Earth, directed upward.
Problem 10 Braking distance and force
1. Firstly, v² = 2as gives a = 400 ÷ (2 × 40) = 5.0 m/s² (opposite to the motion).2. F = ma = 1000 × 5.0 = 5000 N.
Problem 11 Dynamics from a graph
1. Firstly, a = 6.0 ÷ 4.0 = 1.5 m/s², F = 2.0 × 1.5 = 3.0 N.2. Then the velocity is constant, so the net force is zero.
3. a = −6.0 ÷ 2.0 = −3.0 m/s², F = 2.0 × (−3.0) = −6.0 N, opposite to the direction of motion.
Problem 12 Skydiver speeds up
1. Firstly, W = 80 × 9.81 = 785 N downward. a = (785 − 500) ÷ 80 = 3.6 m/s² downward.2. at terminal speed a = 0, so drag equals weight: 785 N.
Problem 13 Two blocks pushed together
1. Firstly, a = 12 ÷ (2.0 + 4.0) = 2.0 m/s².2. Then the 4.0 kg block alone is accelerated by this contact force: F = 4.0 × 2.0 = 8.0 N.
Problem 14 Spot the error
1. Firstly, both forces act on the book. A third-law pair acts on two different objects, and the weight and normal force can differ in size (for example in an elevator).2. Then the partner of the weight is the gravitational force of the book on Earth, directed upward.
Problem 15 Pulling a crate
1. Firstly, a = 120 cos 35° ÷ 25 = 98.3 ÷ 25 = 3.9 m/s².2. N = mg − F sin 35° = 245.3 − 68.8 = 176 N.
Problem 16 Truck and trailer
1. Firstly, a = 5600 ÷ (2000 + 800) = 2.0 m/s².2. Then the coupling accelerates only the trailer: T = 800 × 2.0 = 1600 N.
3. Finally, the tension must accelerate only the trailer, while the 5600 N also accelerates the truck itself.
Problem 17 Loaded lift cable
The total mass is 1500 kg.1. Firstly, T = 1500 × (9.81 + 0.80) = 1.6 × 10⁴ N (15 900 N).
2. a = −1.5 m/s²: T = 1500 × (9.81 − 1.5) = 1.2 × 10⁴ N (12 500 N).
3. T = 1500 × 9.81 = 1.5 × 10⁴ N (14 700 N).
Problem 18 Bob in an accelerating van
1. Firstly, two forces act on the bob: weight mg downward and tension T along the string, up and towards the front. The vertical component of T balances weight and the horizontal component produces the acceleration.2. T cos θ = mg and T sin θ = ma, so tan θ = a ÷ g = 2.5 ÷ 9.81 = 0.255 and θ = 14°.
3. T = m√(g² + a²) = 0.20 × √(96.2 + 6.25) = 0.20 × 10.12 = 2.0 N.
Problem 19 Finding the mass from data
1. Firstly, gradient = (2.01 − 0.51) ÷ (4.0 − 1.0) = 0.50 m/s² per N.2. gradient = 1 ÷ m, so m = 1 ÷ 0.50 = 2.0 kg.
3. Finally, a negative a-intercept means that part of the applied force is needed to overcome a resistive force such as friction before the trolley accelerates.
