Work, motion and forces

Evidence: The passage says "a flat line means the object is not moving." On the graph, segment B is flat: Marcus stays 300 m from home from 100 s to 200 s.

Step by step:

1. Moving means the distance from home is changing.
2. Segment A rises from 0 m to 300 m, segment C rises from 300 m to 800 m, and segment D rises from 800 m to 1,000 m. In all three, the distance changes, so he is moving.
3. In segment B the distance stays at 300 m for 100 seconds. Time passes but distance does not change, so he is at rest (maybe waiting at a traffic light).

The answer is segment B.

Tip: On a distance-time graph, a horizontal line means "stopped." Do not confuse it with a speed-time graph, where a horizontal line means steady speed.

Evidence: The stem gives the astronaut's mass (80 kg) and says the Moon's gravity is "about one-sixth as strong as Earth's."

Step by step:

1. Mass is the amount of matter in an object. Flying to the Moon does not add or remove any of the astronaut's matter, so her mass stays 80 kg.
2. Weight is the force of gravity on that mass (weight = mg). On Earth: 80 \times 9.8 = 784 N.
3. With gravity one-sixth as strong, her weight is about 784 \div 6 \approx 131 N, roughly one-sixth of her Earth weight.
4. Her weight is not zero: the Moon has gravity (that is why astronauts land back on the surface after jumping).

The correct statement is Her mass is still 80 kg, and her weight is about one-sixth of her weight on Earth.

Tip: Mass (kg) is the same everywhere. Weight (N) depends on where you are. If a choice changes the mass, eliminate it.

Evidence: The bus was "moving forward at a steady speed," so the passenger was moving forward with it. The brakes act on the bus, not directly on the passenger.

Step by step:

1. Newton's first law (inertia): an object in motion keeps moving at the same speed and direction unless a net force acts on it.
2. When the brakes slow the bus, the passenger's body keeps moving forward at the old speed for a moment, because nothing has yet acted on it to slow it.
3. Relative to the slowing bus, the passenger seems to be thrown forward. No forward force is actually needed.
4. The forward lurch stops when the floor (friction on feet), a pole or a seat back pushes the passenger backward.

The answer is The passenger's body tends to keep moving forward at the bus's earlier speed (inertia).

Check: The same idea explains why seat belts are needed: they supply the backward force that a passenger's body needs to slow down with the car.

Tip: If an object seems "thrown" with no push, look for inertia. Objects resist changes to their motion; they do not need a force to keep moving.

Evidence: The diagram shows a rope pull of 150 N to the right and friction of 40 N to the left. Gravity (245 N down) and the normal force (245 N up) are equal and opposite. The passage says "Forces in opposite directions are subtracted."

Step by step:

1. Vertical: 245\text{ N up} - 245\text{ N down} = 0. The sled does not move up or down.
2. Horizontal: the forces point in opposite directions, so subtract: F_{\text{net}} = 150\text{ N} - 40\text{ N} = 110\text{ N}
3. The larger force points right, so the net force points right, the direction of the pull.

The net force is 110 N to the right.

Check: If the forces were added (190 N), friction would be helping the pull, but friction always opposes the sliding motion.

Tip: Find the net force one direction at a time. Same direction: add. Opposite directions: subtract, and the net force points the way of the larger force.

Evidence: The table shows Leg 1: "Flew from the warehouse to the house," 1,800 m in 120 s. The passage gives v = \dfrac{d}{t}.

Step by step:

1. Use only Leg 1's numbers: d = 1{,}800 m, t = 120 s.
2. Divide distance by time: v = \frac{1{,}800\text{ m}}{120\text{ s}} = 15\text{ m/s}

The drone flew at 15 m/s during Leg 1.

Check: 15\text{ m/s} \times 120\text{ s} = 1{,}800 m, matching the table. Flipping the fraction (120 ÷ 1,800) gives about 0.067, which would be seconds per meter, not meters per second.

Tip: The units tell you the order: m/s means meters on top, seconds on the bottom.

Evidence: The graph shows that at every force, the 2 kg cart's acceleration is half the 1 kg cart's: at 2 N, 2 m/s² (1 kg) vs. 1 m/s² (2 kg); at 4 N, 4 m/s² vs. 2 m/s². The passage states the law: "the acceleration of an object equals the net force on it divided by its mass."

Step by step:

1. Fix the force and compare: at 4 N, the 1 kg cart accelerates at 4 m/s² and the 2 kg cart at 2 m/s².
2. Doubling the mass (1 kg → 2 kg) halved the acceleration (4 → 2 m/s²). The same pattern holds at 1, 2 and 3 N.
3. Check with a = \dfrac{F}{m}: \dfrac{4}{1} = 4 and \dfrac{4}{2} = 2. ✓
4. Both lines rise, so force does affect acceleration, which rules out a "no effect" choice.

The answer is For the same net force, doubling the cart's mass cut its acceleration in half.

Tip: To compare two groups on one graph, hold one variable steady (pick one x-value) and read both lines at that point.

Evidence: The graph ends at (500 s, 1,000 m): Marcus traveled 1,000 m in 500 s. The passage says average speed is found "by dividing the distance traveled by the time it took."

Step by step:

1. Total distance: he ends 1,000 m from home, starting at 0 m, so 1,000 m.
2. Total time: the trip lasts from 0 s to 500 s, which is 500 s. This includes the 100 s he was stopped in part B.
3. Average speed: v = \frac{1{,}000\text{ m}}{500\text{ s}} = 2\text{ m/s}
4. Check against the segment speeds: A = 3, B = 0, C = 5 and D = 1 m/s. The average falls between them, so 2 m/s is reasonable.

The answer is 2 m/s.

Tip: Never average the segment speeds. Segments last different amounts of time, so you must divide total distance by total time.

Evidence: The diagram shows a push of 50 N to the right and friction of 20 N to the left. The passage says "Forces in opposite directions are subtracted" and that the normal force and gravity "are equal, so they cancel."

Step by step:

1. Vertical: normal force 147 N up and gravity 147 N down. 147 - 147 = 0, so they cancel.
2. Horizontal: 50 N right and 20 N left point in opposite directions, so subtract: 50 - 20 = 30 N.
3. The larger force points right, so the net force points right.

The net force is 30 N to the right, so the box speeds up to the right.

Tip: Net force has a size and a direction. Subtract opposite forces, then give the result the direction of the larger force.

Evidence: The diagram shows the rock weighs 800 N, the load arm is 0.5 m and the effort arm is 2.0 m. The passage gives MA = \dfrac{\text{effort arm length}}{\text{load arm length}} = \dfrac{\text{output force (load)}}{\text{input force (effort)}}.

Step by step:

1. Mechanical advantage: MA = \dfrac{2.0\text{ m}}{0.5\text{ m}} = 4.
2. Rearrange MA = \dfrac{\text{load}}{\text{effort}} to get effort = \dfrac{\text{load}}{MA}.
3. Effort = \dfrac{800\text{ N}}{4} = 200 N.
4. Check: the lever lets her push with a force 4 times smaller, but her end of the bar moves 4 times farther. The work is the same.

The answer is 200 N.

Tip: A machine with MA greater than 1 makes the input force smaller than the load. If your answer is bigger than the load, you multiplied instead of divided.

Evidence: The passage gives the rover's mass as "900 kg," the formula W = mg, and the table lists Mars's g as 3.7 m/s².

Step by step:

1. Use the mass (it is the same everywhere): m = 900 kg.
2. Use Mars's gravitational field strength: g = 3.7\ \text{m/s}^2.
3. Multiply: W = 900 \times 3.7 = 3{,}330\text{ N}

The rover weighs 3,330 N on Mars.

Check: On Earth it would weigh 900 \times 9.8 = 8{,}820 N. Mars's answer is less than half of that, which fits 3.7 being less than half of 9.8.

Tip: Mass (kg) never changes with location; weight (N) does. To get weight, always multiply the mass by the local g.

Evidence: The table gives each mass and speed, and the stem gives p = mv.

Step by step:

1. Hockey puck: 0.16 \times 40 = 6.4 kg m/s.
2. Skater: 70 \times 0.1 = 7.0 kg m/s.
3. Ice-resurfacing machine: 3{,}000 \times 0.002 = 6.0 kg m/s.
4. Stick blade: 0.5 \times 10 = 5.0 kg m/s.
5. The largest product is 7.0 kg m/s.

The skater has the greatest momentum.

Check: The puck is the fastest and the machine is the heaviest, yet the skater wins because momentum depends on both mass and speed together.

Tip: For "which has more momentum" questions, always multiply. A slow, heavy object can beat a fast, light one, and the reverse can also happen.

Evidence: The car "starts from rest" (initial velocity 0 m/s) and reaches 20 m/s, so \Delta v = 20 m/s. The acceleration is 2.5 m/s².

Step by step:

1. Start with a = \dfrac{\Delta v}{t} and solve for t: t = \dfrac{\Delta v}{a}.
2. Substitute: t = \frac{20\text{ m/s}}{2.5\ \text{m/s}^2} = 8\text{ s}

It takes 8 s.

Check: An acceleration of 2.5 m/s² means the car gains 2.5 m/s every second: 2.5, 5, 7.5, 10, 12.5, 15, 17.5, 20 m/s after 8 seconds. ✓

Tip: When the unknown is in the denominator, swap it with the known quantity: t = \Delta v \div a. Check that the units work out: (m/s) ÷ (m/s²) = s.

Evidence: The graph shows part Q as a flat line at 12 m/s that starts at 10 s and ends at 40 s.

Step by step:

1. Velocity during Q: 12 m/s (constant).
2. Elapsed time: 40 - 10 = 30 s. (Use the length of the part, not the clock reading at its end.)
3. Rearrange v = \dfrac{d}{t} to d = vt: d = 12\text{ m/s} \times 30\text{ s} = 360\text{ m}

The bus traveled 360 m during part Q.

Check: On a velocity-time graph, distance equals the area under the line. The area under Q is a rectangle 30 s wide and 12 m/s tall: 30 \times 12 = 360 m. ✓

Tip: Elapsed time = end time − start time. A common mistake is using the clock reading (40 s) as the duration.

Evidence: The passage says "The average speed for a whole trip is the total distance traveled divided by the total time, including any time spent not moving." The table gives distances 1,800 m, 0 m and 1,800 m and times 120 s, 60 s and 180 s.

Step by step:

1. Total distance: 1{,}800 + 0 + 1{,}800 = 3{,}600 m.
2. Total time: 120 + 60 + 180 = 360 s (the hover counts).
3. Divide: v_{\text{avg}} = \frac{3{,}600\text{ m}}{360\text{ s}} = 10\text{ m/s}

The average speed was 10 m/s.

Check: Averaging the leg speeds does not work. Leg 1 was 15 m/s and Leg 3 was 10 m/s, but the drone spent longer on the slower leg, so the true average cannot be the simple average.

Tip: Average speed is never the average of the speeds. Always use total distance ÷ total time.

Evidence: The passage states Newton's third law: "when one object pushes or pulls on a second object, the second object pushes or pulls back on the first with a force of equal size in the opposite direction."

Step by step:

1. The action is: rope pulls on sled, 150 N forward.
2. Swap the two objects: the reaction is sled pulls on rope.
3. Same size, opposite direction: 150 N backward.
4. Friction (40 N) acts on the sled too, but it comes from the snow and is a different size. Gravity and the normal force are vertical and also act on the sled. None of these is the partner of the rope's pull.

The reaction force is the sled pulling backward on the rope with 150 N.

Check: Action-reaction forces never cancel each other, because they act on different objects. That is why the sled can still accelerate.

Tip: To find a reaction force, write A pushes/pulls on B and simply flip it to B pushes/pulls on A, keeping the same size.

Evidence: The stem says the spacecraft is "far out in empty space" and the engine "forces hot exhaust gas out of the back."

Step by step:

1. Newton's third law: when one object pushes on another, the second pushes back with an equal and opposite force.
2. The spacecraft (its engine) pushes the exhaust gas backward.
3. The exhaust gas pushes the spacecraft forward with a force of equal size.
4. That forward force is an unbalanced force on the spacecraft, so it speeds up (Newton's second law).
5. Empty space has no air, so "pushing against the air" cannot be the reason. Rockets actually work best in a vacuum.

The answer is The spacecraft pushes the exhaust backward, and the exhaust pushes the spacecraft forward with an equal force.

Check: You can feel the same effect when you let go of an untied balloon: air rushes out one way and the balloon shoots the other way.

Tip: For any "how does it move forward?" question, find what the object pushes backward. Swimmers push water, walkers push the ground, rockets push exhaust.

Evidence: The table gives Cart A: 2 kg at 3 m/s, and Cart B: 1 kg at 0 m/s. The passage says "the total momentum after the collision equals the total momentum before it," and that the carts "roll away as one unit."

Step by step:

1. Momentum before: Cart A p = 2 \times 3 = 6 kg·m/s; Cart B p = 1 \times 0 = 0. Total = 6 kg·m/s.
2. After the collision, the carts move together with a combined mass of 2 + 1 = 3 kg.
3. Conservation of momentum: 3 \times v = 6, so v = \frac{6\text{ kg}\cdot\text{m/s}}{3\text{ kg}} = 2\text{ m/s (to the right)}
4. Reasonableness: the pair must move slower than Cart A did, because the same momentum is now shared by more mass. 2 m/s < 3 m/s. ✓

The answer is 2 m/s to the right.

Tip: In "stick together" collisions: total momentum before ÷ total mass after = shared velocity. Do not simply average the two speeds.

Evidence: The passage says "A machine can reduce the force needed, but it cannot reduce the work: without friction, the work put in equals the work done on the load." It also gives W = Fd.

Step by step:

1. By hand: force 600 N over 1.0 m: W = 600 \times 1.0 = 600 J.
2. Ramp: effort force = 600 \div 4 = 150 N over 4.0 m: W = 150 \times 4.0 = 600 J.
3. The force is smaller with the ramp, but the distance is larger by the same factor, so the work is the same.

The answer is The work is the same, 600 J, for both methods.

Check: Either way, the machine ends up 1.0 m higher, so the same 600 J of energy is added to it. Energy has to come from somewhere, and the movers supply it.

Tip: Machines change force or distance, not work (ignoring friction). In real life, friction makes the work with a machine slightly more, never less.

Evidence: The graph shows part R starting at 12 m/s at 40 s and ending at 0 m/s at 46 s. The passage says "A negative acceleration means the bus is slowing down."

Step by step:

1. Change in velocity: v_{\text{final}} - v_{\text{initial}} = 0 - 12 = -12 m/s.
2. Time for part R: 46 - 40 = 6 s.
3. Divide: a = \frac{-12\text{ m/s}}{6\text{ s}} = -2.0\ \text{m/s}^2

The acceleration was −2.0 m/s².

Check: The bus loses 2 m/s of speed each second: 12, 10, 8, 6, 4, 2, 0 over 6 seconds. ✓ The negative sign matches the downward slope.

Tip: Always subtract in the order final − initial. When an object slows down, the result is negative.

Evidence: The passage says mass "does not change from place to place" and gives W = mg, with g = 3.7\ \text{m/s}^2 on Mars and 9.8\ \text{m/s}^2 on Earth.

Step by step:

1. Find the mass from the Mars weight: m = \dfrac{W}{g} = \dfrac{74\text{ N}}{3.7\ \text{m/s}^2} = 20 kg.
2. The mass stays 20 kg on Earth.
3. Find the Earth weight: W_{\text{Earth}} = 20 \times 9.8 = 196\text{ N}

The instrument weighs 196 N on Earth.

Check: Earth's gravity is about 9.8 \div 3.7 \approx 2.65 times Mars's, and 74 \times 2.65 \approx 196. The Earth weight should be larger, and it is.

Tip: To move a weight from one world to another, go through the mass: divide by the old g, then multiply by the new g.

Evidence: The diagram shows a 150 N pull to the right and 40 N of friction to the left; the vertical forces (245 N each) cancel. The passage gives the mass as "25 kg" and Newton's second law as F_{\text{net}} = ma.

Step by step:

1. Net force: 150 - 40 = 110 N to the right.
2. Solve for acceleration: a = \dfrac{F_{\text{net}}}{m}.
3. Substitute: a = \frac{110\text{ N}}{25\text{ kg}} = 4.4\ \text{m/s}^2

The sled accelerates at 4.4 m/s² to the right.

Check: 25 \times 4.4 = 110 N, which equals the net force. ✓

Tip: Newton's second law uses the net force. Find it first, then divide by the mass.

Evidence: The passage states Newton's first law: "an object keeps moving at a constant velocity unless a net force acts on it." The diagram shows friction of 40 N acting backward on the sled.

Step by step:

1. Constant velocity means no acceleration, so by F_{\text{net}} = ma, the net force must be zero.
2. Horizontally, only the rope pull and friction act. For them to cancel, they must be equal in size: F_{\text{pull}} - 40\text{ N} = 0 \quad\Rightarrow\quad F_{\text{pull}} = 40\text{ N}
3. A pull of 0 N would leave friction unbalanced, and the sled would slow down and stop. A pull of 150 N would keep it speeding up.

The parent should pull with 40 N.

Check: With 40 N forward and 40 N backward, the forces are balanced, and a moving object with balanced forces keeps moving steadily.

Tip: Balanced forces do not mean not moving. They mean no change in motion. A moving object needs a forward force only to cancel friction.

Evidence: The table gives the load, height and time for each elevator, and the stem gives W = mgh and P = \dfrac{W}{t}.

Step by step:

1. Work, elevator X: 800 \times 9.8 \times 20 = 156{,}800 J.
2. Work, elevator Y: 1{,}000 \times 9.8 \times 15 = 147{,}000 J.
3. Power, X: 156{,}800 \div 16 = 9{,}800 W.
4. Power, Y: 147{,}000 \div 10 = 14{,}700 W.
5. So X does more work, but Y has more power, because Y finishes its job in much less time.

The answer is Elevator X does more work, but elevator Y delivers more power.

Check: Y lifts a heavier load a shorter distance, so the work is close (147,000 J vs 156,800 J), but Y takes only 10 s instead of 16 s. Speed of doing work is what power measures.

Tip: Work is how much energy is transferred; power is how fast. Always compute both before comparing.