Forces and Equilibrium: Lesson Test

Answer: D

Correct answer: newton (N). Newton's second law, force = mass × acceleration, defines the unit: one newton is the force that accelerates 1 kg at 1 m/s², so 1 N = 1 kg·m/s². In equilibrium problems you balance forces such as weights, tensions, and normal forces, so every term must be in newtons. The common trap is the kilogram, which measures mass; to get a weight in newtons, multiply the mass by about 9.8 m/s². Joules (force × distance) and watts (joules per second) build on the newton but measure energy and power.

• A: Kilograms measure mass; kilograms become weight only after multiplying by gravitational acceleration.
• B: A joule measures energy or work, not the magnitude of a force.
• C: A watt measures power, the rate of energy transfer.
• D (correct): A newton is the SI force unit, equivalent to a kilogram meter per second squared.

Answer: D

Correct answer: Velocity. A vector needs both a magnitude and a direction. Velocity, force, acceleration and displacement are vectors, while mass, time, temperature, distance and speed are scalars.

• A: A temperature like 20 degrees has no direction, so it is a scalar.
• B: Mass has only a size, such as 5 kg, so it is a scalar.
• C: An amount of time such as 30 seconds needs no direction, so it is a scalar.
• D (correct): Velocity has both a size and a direction, such as 15 m/s north, so it is a vector.

Answer: A

Correct answer: 60 N. Use W = mg. With m = 6 kg and

the weight is

• A (correct): Weight is mass times g: 6 × 10 = 60 N.
• B: This divides the mass by g instead of multiplying.
• C: This adds an extra factor of 10 to the product of mass and g.
• D: This just repeats the mass number; kilograms must be multiplied by g to get newtons.

Answer: C

Correct answer: It keeps moving in a straight line at a constant speed. Newton's first law: an object with zero net force keeps a constant velocity. A puck that is already moving keeps going straight at the same speed.

• A: Speeding up is an acceleration, which would require a net forward force.
• B: Slowing down is a change in velocity, which would require a net backward force.
• C (correct): With no net force, Newton's first law says the velocity does not change, so speed and direction stay the same.
• D: Objects do not need a force to keep moving; they need a force to change their motion.

Answer: C

Correct answer: 40 N, pointing away from the wall. Newton's third law says forces come in equal and opposite pairs. The skater pushes the wall with 40 N, so the wall pushes the skater with 40 N in the opposite direction, away from the wall.

• A: The pair forces are equal in size, so the wall's push is the full 40 N, not half of it.
• B: Walls and floors can push; this push is a normal force and is what sends the skater backward.
• C (correct): By Newton's third law, the wall pushes back on the skater with a force equal in size and opposite in direction.
• D: The reaction force is opposite to the skater's push, so it points away from the wall, not toward it.

Answer: D

Correct answer: 35 N to the right. Choose right as positive and give the leftward forces negative signs.

The result is positive, so the net force is 35 N to the right.

• A: The size is right, but the larger force points right, so the net force does too.
• B: This subtracts only the 15 N force and forgets the 30 N friction.
• C: This adds all three magnitudes even though two of the forces point the other way.
• D (correct): Taking right as positive: 80 − 30 − 15 = 35 N, and the positive sign means right.

Answer: B

Correct answer: 10 N·m. For a perpendicular push,

Here 40 N × 0.25 m = 10 N·m.

• A: This divides the force by the distance instead of multiplying.
• B (correct): Torque is force times moment arm: 40 × 0.25 = 10 N·m.
• C: This uses 2.5 m instead of 0.25 m, a decimal-place slip.
• D: This ignores the moment arm and reports the force alone.

Answer: B

Correct answer: 30 N. Use Hooke's law, F = kx. With k = 200 N/m and x = 0.15 m,

pulling back toward the natural length.

• A: This uses 1.5 m instead of 0.15 m for the stretch.
• B (correct): Hooke's law gives F = kx = 200 × 0.15 = 30 N.
• C: This divides the spring constant by the stretch instead of multiplying.
• D: This misplaces the decimal point when multiplying 200 by 0.15.

Answer: D

Correct answer: 250 N. A fixed pulley only changes the direction of the pull. Its ideal mechanical advantage is 1, so holding or steadily lifting a 250 N bucket takes a 250 N pull.

• A: Halving the load applies to a movable pulley, not a fixed one.
• B: This divides the load by 4 as if four rope segments held the bucket.
• C: A pulley does not double the force needed; an ideal fixed pulley only changes direction.
• D (correct): A fixed pulley has a mechanical advantage of 1, so the pull equals the load.

Answer: D

Correct answer: Magnetic force. Gravity, magnetic and electric forces act at a distance. Friction, tension and normal forces are contact forces that need surfaces to touch.

• A: A normal force is the push of a surface on an object touching it, so it needs contact.
• B: A rope must be attached to the object to pull on it, so tension is a contact force.
• C: Friction acts only where two surfaces are in contact.
• D (correct): A magnet can pull a steel screw across a gap of air, so magnetic force acts at a distance.

Answer: C

Correct answer: 15 N. Perpendicular forces combine with the Pythagorean theorem.

• A: This subtracts the forces as if they pointed in opposite directions.
• B: This simply adds the magnitudes, which only works when forces point the same way.
• C (correct): The forces are perpendicular, so R = √(9² + 12²) = √225 = 15 N.
• D: This subtracts the squares instead of adding them under the square root.

Answer: A

Correct answer: 2 m/s². First find the net force:

N forward. Then use Newton's second law.

• A (correct): The net force is 120 − 40 = 80 N, and 80 ÷ 40 kg = 2 m/s².
• B: This divides the mass by the net force instead of the net force by the mass.
• C: This adds friction to the push instead of subtracting it: 160 ÷ 40.
• D: This divides the applied push by the mass and ignores friction.

Answer: D

Correct answer: 150 N. A lever balances when

Three times the distance means one-third the force.

• A: This multiplies by both distances instead of dividing by the effort arm.
• B: This ignores the distances; a longer effort arm lets a smaller force balance the load.
• C: This puts the distances in the wrong places, multiplying by 1.2 and dividing by 0.4.
• D (correct): Balancing torques: 450 × 0.4 = F × 1.2, so F = 180 ÷ 1.2 = 150 N.

Answer: B

Correct answer: 120 N. The maximum available friction is

The 120 N push is below that limit, so static friction adjusts to exactly 120 N and the desk stays at rest.

• A: This subtracts the push from the maximum; friction does not equal the leftover amount.
• B (correct): Static friction only grows as large as needed; it exactly matches the 120 N push.
• C: The desk stays still because friction is opposing the push, so friction cannot be zero.
• D: That is the maximum static friction, 0.5 × 400, not the friction actually acting.

Answer: A

Correct answer: 40 N. For a force at angle θ above horizontal,

Here 80 × cos 60° =

• A (correct): The horizontal component is F cos θ = 80 × 0.5 = 40 N.
• B: The full tension is slanted; only part of it acts horizontally.
• C: This uses sine, which gives the vertical component, not the horizontal one.
• D: This divides by cos 60° instead of multiplying, which would make a component bigger than the force.

Answer: B

Correct answer: 720 N. The scale reads the normal force. For upward acceleration,

• A: This is only the extra force needed for the acceleration, 60 × 2, not the full reading.
• B (correct): Accelerating upward, the floor must push more than the weight: 60 × (10 + 2) = 720 N.
• C: This uses g − a, which applies when the acceleration points downward.
• D: This is the true weight, which the scale shows only when the acceleration is zero.

Answer: B

Correct answer: 600 N. Take torques about the left support:

so the right support is 300 N. The supports must total 900 N, so the left support is

• A: This is the torque 900 × 2 in N·m, not a support force.
• B (correct): Torques about the left end give the right support 900 × 2 ÷ 6 = 300 N, so the left carries 900 − 300 = 600 N.
• C: This is the right support's force; the left support is closer to the crate and carries more.
• D: This splits the load evenly, which is only true when the crate is centered.

Answer: C

Correct answer: 750 N. Both the machine and the uniform shelf's weight act at the center. Total load =

By symmetry each bracket carries

• A: This splits only the machine's weight and leaves out the shelf's own 300 N.
• B: That is the total load; two brackets share it, so each carries only half.
• C (correct): The total load is 1,200 + 300 = 1,500 N, centered, so each bracket carries half: 750 N.
• D: This adds the full shelf weight to each bracket instead of splitting it between them.

Answer: A

Correct answer: 250 N. On a frictionless incline, the holding force equals the downhill component of weight.

• A (correct): The force must match the downhill part of the weight: 500 × sin 30° = 250 N.
• B: This uses cos 30°, which gives the part of the weight pressing into the ramp.
• C: A ramp lets you hold a load with less than its full weight; only the downhill component must be balanced.
• D: This divides the weight by sin 30° instead of multiplying.

Answer: B

Correct answer: 200 N. Kinetic friction is

At constant velocity the net force is zero, so the push must also be 200 N.

• A: This sets the push equal to the normal force and forgets the coefficient of friction.
• B (correct): Constant velocity means zero net force, so the push equals kinetic friction: 0.25 × 800 = 200 N.
• C: This divides the normal force by the coefficient instead of multiplying.
• D: Once moving, the crate would still be slowed by friction; a push equal to friction is needed to keep velocity constant.

Answer: D

Correct answer: A bar pushed by two equal, opposite forces on parallel but offset lines. Equilibrium needs both zero net force and zero net torque. Two equal, opposite forces that are not on the same line produce a couple, so the object rotates even though the net force is zero.

• A: Constant velocity in a straight line means zero net force, which is equilibrium.
• B: The cord's tension balances the lamp's weight along the same line, so it is in equilibrium.
• C: Its weight and the normal force balance, and nothing turns it, so it is in static equilibrium.
• D (correct): Offset equal and opposite forces form a couple; the net force is zero but the net torque is not, so the bar starts to spin.

Answer: C

Correct answer: It is not zero and points toward the center of the curve. Velocity includes direction. Going around a curve changes the direction, so there is an acceleration and a net force toward the center of the curve, supplied by tire friction.

• A: An outward net force would push the truck off the curve; the net force must point inward to turn it.
• B: Constant speed is not constant velocity; a change in direction requires a net force.
• C (correct): The truck's direction keeps changing, so it is accelerating; friction from the road supplies a net force toward the center.
• D: A forward net force would speed the truck up, but its speed is not changing.

Answer: A

Correct answer: 2.4 m. Balance requires equal torques on both sides.

• A (correct): Setting torques equal: 600 × 1.0 = 250 × d, so d = 600 ÷ 250 = 2.4 m.
• B: Equal distances balance only equal weights; the lighter child must sit farther out.
• C: This subtracts the parent's 1.0 m from the correct distance, but distance is measured from the pivot.
• D: This divides the weights the wrong way, putting the lighter child closer than the heavier parent.

Answer: A

Correct answer: 5 cm. At rest, the spring force equals the weight: W = mg =

From Hooke's law,

m = 5 cm.

• A (correct): The spring force must equal the 20 N weight, so x = 20 ÷ 400 = 0.05 m, or 5 cm.
• B: This uses the 2 kg mass as if it were a force instead of the 20 N weight.
• C: This divides the spring constant by the weight and reads the result as centimeters.
• D: This makes a decimal slip when converting 0.05 m to centimeters.

Answer: C

Correct answer: It accelerates north, with a net force of 30 N. Treat each direction separately. East-west:

North-south:

N toward the north. Since one sum is not zero, the ring accelerates north.

• A: Both the horizontal and the vertical sums must be zero; the vertical sum is not.
• B: This adds all four magnitudes, ignoring that opposite forces subtract.
• C (correct): Horizontal forces cancel (50 − 50 = 0), but vertically 70 − 40 = 30 N is left over toward the north.
• D: The larger vertical pull is the 70 N northward one, so the leftover force points north.

Answer: C

Correct answer: 300 N. Only the vertical parts of the two tensions support the weight:

• A: Half the weight would be correct only if the ropes hung straight up.
• B: This uses cos 30° instead of sin 30°, but only the vertical parts of the tensions hold up the sign.
• C (correct): T = W ÷ (2 sin θ) = 300 ÷ (2 × 0.5) = 300 N.
• D: This divides the weight by sin 30° but forgets that two ropes share the load.

Answer: A

Correct answer: 8. Ideal lever mechanical advantage is the effort arm divided by the load arm:

A push at the handle produces eight times as much force at the load.

• A (correct): For an ideal lever, MA = effort arm ÷ load arm = 1.6 ÷ 0.2 = 8.
• B: This subtracts the arms instead of dividing them.
• C: This adds the arms instead of dividing them.
• D: This divides the load arm by the effort arm, which is upside down.

Answer: B

Correct answer: Place the heaviest boxes on the bottom. Wide bases and low centers of gravity resist tipping. Loading the heaviest items low on the cart lowers the center of gravity.

• A: Heavy items up at the handle raise the center of gravity rather than lowering it.
• B (correct): Heavy items low keep the center of gravity low, which makes the cart more stable.
• C: A tall, narrow load raises the center of gravity and narrows the base, which reduces stability.
• D: Heavy items on top raise the center of gravity, making tipping easier.

Answer: C

Correct answer: 70 N. With no vertical acceleration, the upward normal force must equal the total downward force:

• A: This subtracts the push, which would be right only if the worker were lifting up.
• B: This counts only the worker's push and leaves out the box's weight.
• C (correct): The floor must balance both the 50 N weight and the 20 N downward push: 50 + 20 = 70 N.
• D: The normal force equals the weight only when no other vertical forces act.

Answer: C

Correct answer: It tips over, once the push reaches 100 N. Tipping push: push ×

so 100 N. Sliding push:

The cabinet reaches the 100 N tipping threshold first, so it tips.

• A: Sliding does need 120 N, but the 100 N tipping threshold is reached before that.
• B: This divides by the half-width and multiplies by the height, reversing the torque balance.
• C (correct): Tipping needs push × 1.0 = 400 × 0.25, so 100 N; sliding needs 0.3 × 400 = 120 N, and the smaller threshold is reached first.
• D: 100 N is the tipping threshold; at 100 N friction can still hold the cabinet in place.

Answer: A

Correct answer: 12 N·m. Only the part of the force perpendicular to the handle turns the bolt.

• A (correct): τ = F d sin θ = 60 × 0.4 × 0.5 = 12 N·m.
• B: This uses cos 30° instead of sin 30° for the angle between the force and the handle.
• C: This divides the force by the distance instead of multiplying.
• D: This would be the torque for a pull at right angles; it ignores the 30° angle.

Answer: D

Correct answer: No; it needs about 85 N of friction but at most about 82 N is available. Compare tan θ with the coefficient: tan 25° ≈ 0.47 is greater than 0.45, so the block slides. In forces: needed friction = 200 × sin 25° ≈ 85 N, while the maximum is 0.45 × 200 × cos 25° ≈ 82 N.

• A: Only the downhill component of weight, about 85 N, must be balanced by friction.
• B: This uses the full weight as the normal force; on a slope the normal force is only W cos θ.
• C: The normal force is perpendicular to the slope and cannot hold the block against the downhill pull.
• D (correct): Needed friction is 200 sin 25° ≈ 85 N; the limit is 0.45 × 200 cos 25° ≈ 82 N, so friction falls short and the block slides.

Answer: B

Correct answer: About 416 N. Only the rope supports the weight vertically:

so T = 480 N. The rope's horizontal pull toward the wall is 480 × cos 30° ≈ 416 N, so the strut pushes outward with about 416 N.

• A: This assumes the strut's push equals the weight; the strut balances the rope's horizontal pull instead.
• B (correct): The rope's vertical part holds the light: T × 0.5 = 240, so T = 480 N; its horizontal part, 480 × 0.87 ≈ 416 N, is balanced by the strut.
• C: That is the rope tension, not its horizontal component balanced by the strut.
• D: This multiplies the weight by tan 30° instead of dividing by it.

Answer: A

Correct answer: 900 N. The acceleration points opposite the velocity because the elevator is slowing, so it points upward. Then

• A (correct): Slowing a downward motion means the acceleration points up, so N = 75 × (10 + 2) = 900 N.
• B: This is the true weight, which the scale reads only at constant speed.
• C: This counts only the 75 × 2 needed for the acceleration, not the full normal force.
• D: This uses g − a, but slowing down while moving down is an upward acceleration.

Answer: B

Correct answer: 300 N of pull and 6 m of rope. Three supporting segments each carry the same tension: 3T = 900, so T = 300 N. To raise the load 2 m, each of the three segments shortens 2 m, so 6 m of rope must be pulled.

• A: This treats the system like a single fixed pulley with no mechanical advantage.
• B (correct): Three segments share the load, so 900 ÷ 3 = 300 N, and each must shorten 2 m, so 3 × 2 = 6 m of rope.
• C: The force is right, but a smaller force always costs more rope distance.
• D: This treats the system as having only two supporting segments.

Answer: A

Correct answer: 950 N. Take torques about the left support, with the beam's weight acting at its 4 m midpoint.

So R(right) = 450 N, and

• A (correct): Torques about the left: R(right) × 8 = 400 × 4 + 1,000 × 2, so R(right) = 450 N; then R(left) = 1,400 − 450 = 950 N.
• B: This finds the load's share at the left support but leaves out the beam's own 400 N weight.
• C: That is the right support's force; the left support is nearer the load and carries more.
• D: This splits the total 1,400 N evenly, ignoring that the load is off-center.

Answer: C

Correct answer: 25 N, pointing between south and west. The resultant of the perpendicular pulls is

It points between north and east, so the equilibrant is 25 N pointing the opposite way, between south and west.

• A: This subtracts the forces as if they acted along the same line.
• B: This adds the magnitudes; perpendicular forces combine by the Pythagorean theorem.
• C (correct): The resultant is √(15² + 20²) = 25 N toward the northeast, so the balancing force is 25 N in the opposite direction.
• D: That is the resultant itself; adding it would double the unbalanced force instead of cancelling it.

Answer: D

Correct answer: 2.6 m/s².

The static limit is

which 280 N exceeds, so the crate slides. Kinetic friction is

the net force is

• A: This divides the push by the mass and ignores friction.
• B: This uses static friction (250 N) after the crate is already sliding; kinetic friction takes over.
• C: The static limit is 0.5 × 500 = 250 N, and 280 N exceeds it, so the crate does move.
• D (correct): 280 N beats the 250 N static limit, so the crate slides; kinetic friction is 150 N, net force 130 N, and 130 ÷ 50 = 2.6 m/s².

Answer: B

Correct answer: The toolbox's gravitational pull upward on Earth. A third-law pair involves the same two objects acting on each other. The weight is Earth pulling on the toolbox, so its partner is the toolbox pulling on Earth. The bench's upward push balances the weight but is not its pair.

• A: This is the partner of the bench's normal force, not of gravity.
• B (correct): Third-law partners act between the same two objects; Earth pulls the toolbox down, so the toolbox pulls Earth up.
• C: This force balances the weight, but it comes from a different interaction and acts on the same object, so it is not the pair.
• D: This acts on the bench and involves the floor, not Earth's gravitational pull on the toolbox.

Answer: D

Correct answer: Net force 0 N; net torque 10 N·m. Force test: 25 N up and 25 N down cancel, so the net force is zero. Torque test: both hands turn the wheel the same way.

• A: The forces form a couple; their torques add rather than cancel.
• B: This counts the torque from only one hand.
• C: Opposite forces subtract, so the net force is zero, not their sum.
• D (correct): Up and down forces cancel, but both turn the wheel the same way: 2 × 25 × 0.2 = 10 N·m.

Answer: D

Correct answer: About 57 N. Split the pull: horizontal part = 100 × 0.87 ≈ 87 N, vertical part =

N upward. The upward part lifts some weight off the snow, so

N and kinetic friction =

Net horizontal force ≈

N forward.

• A: This sets the normal force equal to the full 200 N weight, giving 40 N of friction; the rope's upward part reduces the normal force.
• B: This uses the whole 100 N as if it were horizontal and also takes friction from the full weight; only the horizontal component pulls the sled forward.
• C: This is only the horizontal part of the pull; friction still acts backward and must be subtracted.
• D (correct): The horizontal pull of about 87 N minus the 30 N of friction, with friction based on the reduced normal force of 150 N.

Answer: D

Correct answer: 75 N. By Hooke's law, force is proportional to stretch. The spring constant is k = 45 N ÷ 0.03 m = 1,500 N/m.

For a 0.05 m stretch,

As a ratio check:

• A: This flips the ratio (45 × 3/5); a larger stretch needs a larger force, not a smaller one.
• B: This adds 2 N for the extra 2 cm instead of scaling the force in proportion to the stretch.
• C: This multiplies 45 by 5 as if 45 N produced only 1 cm of stretch.
• D (correct): Force grows in proportion to stretch: 5 cm is 5/3 of 3 cm, and 45 × 5/3 = 75 N.

Answer: A

Correct answer: 56 N·m. Convert the moment arm to meters first: 35 cm = 0.35 m.

Torque = force × perpendicular distance =

N·m.

• A (correct): Force times the moment arm in meters: 160 × 0.35 = 56 N·m.
• B: This uses 35 cm as if it were 35 m; centimeters must be converted to meters before computing torque in N·m.
• C: This divides the force by the distance; torque is force multiplied by the moment arm.
• D: This treats 35 cm as 3.5 m, a factor-of-ten slip in the unit conversion.

Answer: B

Correct answer: About 520 N. On a slope the weight splits into a part along the ramp (W sin θ) and a part pressing into the ramp (W cos θ). With no other forces perpendicular to the surface, the normal force balances the part pressing in.

N = 600 × cos 30° ≈ 600 × 0.87 ≈ 520 N.

• A: This assumes the normal force always equals the weight, which is true only on level ground.
• B (correct): The normal force balances the weight component into the ramp, W cos θ ≈ 600 × 0.87.
• C: This is W sin θ, the part of the weight that tries to pull the crate down the ramp, not the part pressing into it.
• D: This divides the weight by cos 30° instead of multiplying; the normal force on a ramp is less than the weight.

Answer: D

Correct answer: 50 N, pointing between south and west. Combine the existing forces by direction. East-west:

N east. North-south: 40 N north. These are perpendicular, so the resultant is

pointing between north and east.

The balancing force must be equal in size and opposite in direction: 50 N pointing between south and west.

• A: This is the resultant of the three ropes; the fourth rope must cancel it, so it points the opposite way.
• B: This adds 30 N and 40 N directly; perpendicular forces combine with the Pythagorean theorem.
• C: This adds every magnitude, ignoring that the east and west pulls partly cancel and that north is at a right angle.
• D (correct): The other ropes add to 50 N toward the northeast, so the fourth rope must pull 50 N the opposite way.

Answer: B

Correct answer: Net torque is 0; drag and friction supply 8 N·m against the motor. A steady spin rate means no change in rotation, so the wheel is in rotational equilibrium and the net torque is zero.

The only way the 8 N·m motor torque can be cancelled is by friction and air drag supplying 8 N·m in the opposite direction.

• A: A net torque of 8 N·m would make the wheel spin faster and faster, not hold a steady speed.
• B (correct): Constant rotation means zero net torque, so the resisting torque must exactly match the motor's 8 N·m.
• C: Any leftover torque would change the spin rate; a steady speed requires the torques to balance exactly.
• D: If nothing resisted it, the motor's 8 N·m would be unbalanced and the wheel would speed up.

Answer: B

Correct answer: 500 N. Only the vertical parts of the two tensions hold up the weight, so 2T sin θ = W.

• A: This simply splits the weight in half, which is right only for vertical cables; angled cables must pull harder.
• B (correct): Solving 2T sin 60° = 870 gives T = 870 ÷ 1.74 = 500 N.
• C: This uses cos 60° in place of sin 60°; the vertical part of a cable's pull uses the sine of its angle above horizontal.
• D: This divides the weight by sin 60° but forgets that two cables share the load.

Answer: C

Correct answer: 825 N. Take torques about the left support so its unknown force drops out. The plank's weight acts at its center, 3 m from the left end; the worker is 6 m out.

This is more than the 700 N total load, so the left end must actually be held down with 125 N.

• A: This counts only the worker's torque (400 × 6 ÷ 4) and leaves out the plank's own 300 N weight at its center.
• B: This assumes the support carries the total load; with the worker on the overhang, the support near him carries more than the total.
• C (correct): Torques about the left support: (300 × 3 + 400 × 6) ÷ 4 = 825 N.
• D: This splits the 700 N total load evenly between the supports, which only works for a centered load between them.

Answer: A

Correct answer: Straight down. After the ball leaves the hand, the only force on it is gravity, which pulls straight down. The ball keeps rising only because of the upward velocity it already has; the downward force is what slows it.

• A (correct): Gravity is the only force after release, and it points down even while the ball is still moving up.
• B: This confuses the direction of motion with the direction of force; no upward force acts once the ball leaves the hand.
• C: Gravity acts the whole time; without it the ball would never slow down and turn around.
• D: The force is gravity alone throughout the flight; only the velocity changes direction at the top.

Answer: C

Correct answer: 120 kg. Mass is the amount of matter in an object and does not change with location. The crate's weight drops to about one-sixth on the Moon, but its mass stays 120 kg.

• A: This divides by six as if mass changed; it is the weight that becomes about one-sixth as large.
• B: This multiplies by six; neither mass nor weight increases on the Moon.
• C (correct): Mass does not depend on location; only the weight changes with the local gravity.
• D: This is roughly the crate's weight on Earth in newtons, which is a force, not a mass.