30 multiple-choice + 2 free-response = 32 questions · First Law · F = ma · weight & elevators · Third Law · free-body diagrams · friction · inclines & pulleys
Closed book, calculator allowed, g = 9.81 m/s². Sketch a free-body diagram before each force problem. Aim for about 40 minutes. Answer everything, then press Submit at the bottom to see your score, the correct answers and an explanation for each one.
Part A — Multiple choice
1First Law
A hockey puck slides across frictionless ice at a steady 4 m/s. What keeps it moving?
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B) Nothing: no force is needed to keep it moving (inertia). By the First Law an object keeps its velocity unless a net force acts. No forward force is needed; a force would only be needed to CHANGE its motion.
Answer: B) Nothing: no force is needed to keep it moving (inertia). By the First Law an object keeps its velocity unless a net force acts. No forward force is needed; a force would only be needed to CHANGE its motion.
2First Law
A car travels at a constant 25 m/s along a straight highway. What is the net force on it?
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A) Zero. Constant velocity means a = 0, so F_net = 0. The forward push from the road exactly balances air resistance and rolling friction.
Answer: A) Zero. Constant velocity means a = 0, so F_net = 0. The forward push from the road exactly balances air resistance and rolling friction.
3First Law
Which object has the MOST inertia?
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C) A 70 kg student. Inertia is measured by mass. 70 kg is the largest mass (1000 g is only 1 kg).
Answer: C) A 70 kg student. Inertia is measured by mass. 70 kg is the largest mass (1000 g is only 1 kg).
4First Law
A stone whirled in a horizontal circle on a string is moving due north when the string breaks. What path does it follow (seen from above)?
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D) A straight line due north (along the tangent). Once the string force is gone there is no horizontal force, so by the First Law the stone keeps the velocity it had: straight north along the tangent. There is no outward 'centrifugal' force.
Answer: D) A straight line due north (along the tangent). Once the string force is gone there is no horizontal force, so by the First Law the stone keeps the velocity it had: straight north along the tangent. There is no outward 'centrifugal' force.
5Second Law
A net force of 24 N acts on a 6.0 kg cart. What is its acceleration?
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C) 4.0 m/s². a = F_net/m = 24/6.0 = 4.0 m/s².
Answer: C) 4.0 m/s². a = F_net/m = 24/6.0 = 4.0 m/s².
6Second Law
The same net force acts on an object of mass m and on one of mass 3m. The acceleration of the 3m object is…
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B) one-third as large. a = F/m. Tripling the mass with the same force gives one-third the acceleration.
Answer: B) one-third as large. a = F/m. Tripling the mass with the same force gives one-third the acceleration.
7Second Law
A 1500 kg car speeds up uniformly from rest to 20 m/s in 10 s. What net force acts on it?
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D) 3000 N. a = (20 − 0)/10 = 2.0 m/s², so F = ma = 1500 × 2.0 = 3000 N.
Answer: D) 3000 N. a = (20 − 0)/10 = 2.0 m/s², so F = ma = 1500 × 2.0 = 3000 N.
8Second Law
A 10 kg box on a frictionless floor is pushed with 50 N to the right and 20 N to the left. What is its acceleration?
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A) 3.0 m/s² to the right. F_net = 50 − 20 = 30 N to the right, so a = 30/10 = 3.0 m/s² to the right.
Answer: A) 3.0 m/s² to the right. F_net = 50 − 20 = 30 N to the right, so a = 30/10 = 3.0 m/s² to the right.
9Second Law
A car moving east is braking. In which direction is the net force on it?
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B) West. Slowing down while moving east means the acceleration points west. F_net always points the same way as a.
Answer: B) West. Slowing down while moving east means the acceleration points west. F_net always points the same way as a.
10Second Law
Which unit is the same as one newton?
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C) kg·m/s². From F = ma: 1 N = 1 kg × 1 m/s² = 1 kg·m/s². (kg·m/s is the unit of momentum.)
Answer: C) kg·m/s². From F = ma: 1 N = 1 kg × 1 m/s² = 1 kg·m/s². (kg·m/s is the unit of momentum.)
11Weight & Apparent Weight
An astronaut has a mass of 80 kg on Earth. What is her mass on the Moon (g = 1.62 m/s²)?
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D) 80 kg. Mass is the amount of matter and doesn't depend on location. Only her weight changes (to 80 × 1.62 = 130 N).
Answer: D) 80 kg. Mass is the amount of matter and doesn't depend on location. Only her weight changes (to 80 × 1.62 = 130 N).
12Weight & Apparent Weight
What is the weight of a 5.0 kg bag of flour on Earth?
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C) 49 N. W = mg = 5.0 × 9.81 = 49 N.
Answer: C) 49 N. W = mg = 5.0 × 9.81 = 49 N.
13Weight & Apparent Weight
A 70 kg person stands on a scale in an elevator accelerating upward at 1.5 m/s². What does the scale read?
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B) 792 N. N − mg = ma, so N = m(g + a) = 70 × (9.81 + 1.5) = 70 × 11.31 = 792 N.
Answer: B) 792 N. N − mg = ma, so N = m(g + a) = 70 × (9.81 + 1.5) = 70 × 11.31 = 792 N.
14Weight & Apparent Weight
In which case does a scale in an elevator read LESS than your true weight?
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D) Moving up and slowing down. The scale reads less when the acceleration points down. Moving up and slowing down means a is downward. (Moving down and slowing down means a is UP, so it reads more.)
Answer: D) Moving up and slowing down. The scale reads less when the acceleration points down. Moving up and slowing down means a is downward. (Moving down and slowing down means a is UP, so it reads more.)
15Weight & Apparent Weight
A skydiver has reached terminal velocity. The air resistance on her is…
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A) equal to her weight. Terminal velocity is constant velocity, so a = 0 and F_net = 0: drag balances weight exactly.
Answer: A) equal to her weight. Terminal velocity is constant velocity, so a = 0 and F_net = 0: drag balances weight exactly.
16Third Law
Earth pulls a falling apple down with a force of 2 N. What is the reaction force?
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B) The apple pulls Earth up with 2 N. Swap A and B: 'Earth pulls apple' becomes 'apple pulls Earth'. The pair is the same type of force (gravity), equal and opposite.
Answer: B) The apple pulls Earth up with 2 N. Swap A and B: 'Earth pulls apple' becomes 'apple pulls Earth'. The pair is the same type of force (gravity), equal and opposite.
17Third Law
A book rests on a table. How are the table's upward push (N) and the book's weight (W) related?
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A) Equal because the book isn't accelerating (First Law); they are NOT a third-law pair. N and W both act on the book and are different types of force, so they can't be a third-law pair. They are equal only because a = 0.
Answer: A) Equal because the book isn't accelerating (First Law); they are NOT a third-law pair. N and W both act on the book and are different types of force, so they can't be a third-law pair. They are equal only because a = 0.
18Third Law
A 50 kg skater pushes a 100 kg skater with a force of 60 N. How big is the force on the 50 kg skater?
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D) 60 N. By the Third Law, the 100 kg skater pushes back on the 50 kg skater with exactly 60 N.
Answer: D) 60 N. By the Third Law, the 100 kg skater pushes back on the 50 kg skater with exactly 60 N.
19Third Law
Two skaters, 50 kg and 100 kg, push off each other. Compared with the 100 kg skater, the 50 kg skater's acceleration is…
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C) twice as large. They feel equal forces (Third Law), and a = F/m, so half the mass means twice the acceleration.
Answer: C) twice as large. They feel equal forces (Third Law), and a = F/m, so half the mass means twice the acceleration.
20Third Law
Which force actually makes a car accelerate forward on a road?
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A) The road pushing forward on the tires (static friction). The tires push back on the road, and by the Third Law the road pushes the tires forward. Only that outside force (on the car) accelerates it.
Answer: A) The road pushing forward on the tires (static friction). The tires push back on the road, and by the Third Law the road pushes the tires forward. Only that outside force (on the car) accelerates it.
21Free-Body Diagrams
A ball has been thrown straight up and is still rising. Ignoring air resistance, which is its correct free-body diagram?
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C) C. After release nothing is pushing it up; the throw ended when it left the hand. The only force is its weight, downward, even while it rises (it is slowing down).
Answer: C) C. After release nothing is pushing it up; the throw ended when it left the hand. The only force is its weight, downward, even while it rises (it is slowing down).
22Free-Body Diagrams
A block slides DOWN a rough incline. Which is its correct free-body diagram?
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D) D. Weight straight down, normal force perpendicular out of the surface, and kinetic friction up the slope (opposing the sliding). A has friction the wrong way, B has no normal force, and C adds a non-existent 'force down the slope'.
Answer: D) D. Weight straight down, normal force perpendicular out of the surface, and kinetic friction up the slope (opposing the sliding). A has friction the wrong way, B has no normal force, and C adds a non-existent 'force down the slope'.
23Free-Body Diagrams
How many forces act on a book resting on a level table?
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C) 2. Weight (Earth pulls it down) and the normal force (table pushes it up). No friction acts because nothing tries to slide it.
Answer: C) 2. Weight (Earth pulls it down) and the normal force (table pushes it up). No friction acts because nothing tries to slide it.
24Free-Body Diagrams
A crate is pulled across a rough floor by a horizontal rope at constant velocity. Which forces act on the crate?
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B) Weight, normal force, tension, friction. Earth (weight), floor (normal and friction) and rope (tension). There is no 'force of motion'; with constant velocity, tension = friction and N = W.
Answer: B) Weight, normal force, tension, friction. Earth (weight), floor (normal and friction) and rope (tension). There is no 'force of motion'; with constant velocity, tension = friction and N = W.
25Friction
A 20 kg box sits on a floor with μs = 0.50 and μk = 0.40. What is the minimum horizontal push that starts it moving?
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A) 98 N. N = mg = 20 × 9.81 = 196.2 N. f_s,max = μs·N = 0.50 × 196.2 = 98 N.
Answer: A) 98 N. N = mg = 20 × 9.81 = 196.2 N. f_s,max = μs·N = 0.50 × 196.2 = 98 N.
26Friction
Once the box from the previous question (20 kg, μk = 0.40) is sliding, what push keeps it moving at constant velocity?
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D) 78 N. Constant velocity means F_net = 0, so the push equals kinetic friction: μk·N = 0.40 × 196.2 = 78 N.
Answer: D) 78 N. Constant velocity means F_net = 0, so the push equals kinetic friction: μk·N = 0.40 × 196.2 = 78 N.
27Friction
You push a box with 40 N and it doesn't move. Its maximum static friction is 90 N. What friction force acts on it?
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C) 40 N. Static friction adjusts to match the push, up to its maximum. The box is at rest, so F_net = 0 and friction = 40 N.
Answer: C) 40 N. Static friction adjusts to match the push, up to its maximum. The box is at rest, so F_net = 0 and friction = 40 N.
28Friction
Why is it easier to keep a heavy box sliding than to start it moving?
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B) μk is usually smaller than μs. Kinetic friction μk·N is usually less than the maximum static friction μs·N for the same surfaces.
Answer: B) μk is usually smaller than μs. Kinetic friction μk·N is usually less than the maximum static friction μs·N for the same surfaces.
29Inclines & Pulleys
A block slides down a frictionless 30° incline. What is its acceleration?
Show answer
D) 4.9 m/s². a = g sinθ = 9.81 × sin30° = 9.81 × 0.5 = 4.9 m/s². (8.5 m/s² would be g cosθ.)
Answer: D) 4.9 m/s². a = g sinθ = 9.81 × sin30° = 9.81 × 0.5 = 4.9 m/s². (8.5 m/s² would be g cosθ.)
30Inclines & Pulleys
Masses of 4.0 kg and 6.0 kg hang over a frictionless pulley (an Atwood machine). What is their acceleration?
A 12 kg box is pulled across a level floor by a horizontal rope with 70 N of tension. μk = 0.35. (a) Draw its free-body diagram. (b) Find its acceleration. (c) Starting from rest, how far does it move in 3.0 s?
Show worked solution
(a) Four forces: weight mg down, normal force N up (from the floor), tension T = 70 N forward (along the rope), kinetic friction f_k backward.
Vertical: a = 0, so N = mg = 12 × 9.81 = 117.7 N.
Friction: f_k = μk·N = 0.35 × 117.7 = 41.2 N.
(b) Horizontal, forward positive: F_net = 70 − 41.2 = 28.8 N, so a = 28.8/12 = 2.4 m/s².
(c) s = ut + ½at² = 0 + ½ × 2.4 × 3.0² = 10.8 m.
32Inclines & Pulleys · free response
A 3.0 kg block on a frictionless table is tied over a pulley at the edge to a 1.0 kg hanging mass. Find (a) the acceleration and (b) the tension. (c) Explain why the tension is less than the hanging weight (9.81 N).
Show worked solution
Block, toward the pulley positive: T = 3.0a …(1). Hanging mass, down positive: 1.0 × 9.81 − T = 1.0a …(2).
(a) Add (1) and (2): 9.81 = 4.0a, so a = 2.45 m/s².
(b) From (1): T = 3.0 × 2.45 = 7.36 N. Check with (2): 9.81 − 7.36 = 2.45 = 1.0 × 2.45 ✓
(c) The hanging mass accelerates downward, so the net force on it must point down: mg − T = ma > 0, so T < mg. If T were 9.81 N the hanging mass could not accelerate.
