Motion and forces — Intermediate

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Level 2: Intermediate

Explain, apply, analyse, sequence — connect ideas, read patterns, use rules in new contexts (bridge to Year 11).

1. A net horizontal force of 450 N produces acceleration 2.5 m/s² on a crate. The crate’s mass is:

   A)1125 kgB)180 kgC)225 kgD)18 kg

2. You push a heavy shopping trolley and it barely accelerates. Why is $F_{\text{net}} = ma$ still consistent?

   A)Your push counts as the net force, so tiny acceleration must mean the trolley’s effective mass is negligible.B)Small acceleration shows $F_{\text{net}}$ stays larger than $ma$ until the wheels break free and roll.C)Friction ceases once you apply a push, so the net force on the trolley equals your push alone.D)Friction and other forces can nearly cancel your push, leaving a small unbalanced force and small acceleration.

3. A book rests on a table. Weight pulls the book down and the table exerts an upward normal force. For a stationary book, these two forces on the book:

   A)Are a Newton’s-third-law pair, because gravity and the table push are equal and opposite interaction partners.B)Balance on the same object so the net force on the book is zero while it remains stationary.C)Add vectorially to a net force equal to weight plus normal combined, producing a downward resultant on the book.D)Act on different bodies by definition, which prevents the book from sitting in static equilibrium on the table.

4. Why do action–reaction partners not cancel when you find the net force on one object?

   A)They automatically differ in size whenever the contact surfaces are horizontal rather than vertical.B)Only one partner should appear on a free-body diagram, so the other partner is dropped by convention alone.C)They act on different bodies, and net force sums only the forces acting on the single object you chose.D)They always align in the same direction on the two objects, so they add instead of opposing each other.

5. A 2.0 kg object accelerates at 6.0 m/s². The net force is:

   A)12 NB)8 NC)3 ND)0.33 N

6. A 5.0 kg block slides at constant speed on a rough horizontal floor. The horizontal net force on the block is best described as:

   A)Equal to the block’s weightB)Equal to kinetic friction only, always larger than the applied pushC)Zero if the speed is truly constant (horizontal forces balance)D)Non-zero because moving objects always have net force in the direction of motion

7. Two forces on one object: 12 N east and 5 N west on the same line. The magnitude of the net force is:

   A)17 NB)7 NC)60 ND)2.4 N

8. A rocket expels gas backward. The rocket accelerates forward partly because:

   A)The exhaust pushes Earth, and Earth immediately pushes the rocket forward through the launch structure.B)Both third-law partners act on the rocket together, so they sum internally and raise the rocket’s net force.C)Mass falls so quickly that $F=ma$ demands acceleration even when thrust from the engines drops to zero.D)The rocket pushes on the exhaust backward, and the exhaust pushes the rocket forward as the paired force.

9. On Earth, weight refers most closely to:

   A)The gravitational force on the object, often written $W = mg$ near the surface.B)The normal support force from a horizontal surface when the object rests on it.C)The object’s mass in kilograms treated as if it were a force on the object.D)The net force on the object whenever it is at rest, with no other forces drawn.

10. A book sits at rest on a table. Earth pulls downward on the book gravitationally. One Newton’s-third-law partner to that force on Earth is:

    A)The table’s upward normal force on the bookB)The upward gravitational pull of the book on EarthC)Static friction between the book and the tableD)The combined weight of the book and table acting on the floor

11. A 4.0 kg cart has a horizontal net force of 16 N on a smooth runway. Its acceleration is:

    A)64 m/s²B)12 m/s²C)4 m/s²D)0.25 m/s²

12. A car travels around a curve at constant speed. Why is this not uniform velocity in the physics sense?

    A)Speed is not constantB)Net force must be zero on a curveC)Acceleration is always zero on a circular pathD)Velocity changes because direction changes, even when speed is steady

13. Two students pull a rope: 90 N east and 40 N west along the same line. The net force on the rope is:

    A)130 NB)50 NC)3600 ND)0 N because tension cancels internally

14. Two students pull a rope in opposite directions along the same line with pulls of equal magnitude. The rope does not move. The net force on the rope is:

    A)Equal to one student’s pull only, because the other pull is taken to act on the surroundings.B)Equal to the vector sum that doubles one pull in the forward direction along the rope.C)Non-zero whenever the rope is horizontal, because weight adds a component along the rope.D)Zero, because equal-magnitude opposite pulls along one line cancel on the rope itself.

15. Before a crate slips, static friction can:

    A)Match an applied horizontal push up to a maximum value; kinetic friction often governs motion once sliding begins.B)Never grow as large as the applied horizontal force, so sliding must begin the instant you start pushing.C)Act only vertically between stacked surfaces, not along the direction you are trying to slide the crate.D)Be neglected whenever the surface looks smooth, because visible smoothness means friction is negligible.

16. A parachute opens before landing mainly because extra air resistance can:

    A)Remove all downward motion in an instant so the body stops without any gradual slowing phase.B)Shorten the remaining stopping time so peak forces on the body become larger before touchdown.C)Increase drag so slowing happens over more time and distance, which tends to lower peak forces on the body.D)Cancel weight completely so vertical acceleration becomes zero for the rest of the descent.

17. A skydiver at terminal speed falls at constant velocity. The net force on the skydiver is best described as:

    A)Equal to weight plus dragB)Zero — drag balances weight at that speedC)Always $mg$ downward regardless of speedD)Larger than weight because air pushes down

18. You push a stack of books horizontally at constant speed across a desk. The horizontal net force on the stack is:

    A)Equal to your push alone, because friction is not a real force once the books are sliding steadily.B)Equal to the stack’s weight, since weight is the main horizontal load when you slide objects on a desk.C)Non-zero in the direction of motion, because moving objects always need a net force pointing along velocity.D)Zero if speed is truly constant, meaning your horizontal push balances friction and there is no unbalanced force.

19. A person steps forward by pushing backward on the ground. The ground pushes the person forward because:

    A)The forward push on the person is the reaction to the backward push on the ground, and it can shape net force on the person.B)Both third-law partners act on the person at once, so they add on the same body and double the forward force.C)Friction on the foot always points forward even when the foot is clearly pushing the ground backward.D)The ground is treated as inertial, so by definition it cannot exert any horizontal reaction force on the person.

20. A heavy truck and a light truck are slowed by the same-size net horizontal braking force. Ignoring other differences, the heavier truck usually has:

    A)Larger acceleration magnitude than the light truckB)Smaller acceleration magnitude than the light truckC)The same acceleration magnitude regardless of massD)Zero acceleration until the tyres stop slipping

21. A student on a low-friction skateboard throws a medicine ball straight forward. The student tends to roll backward mainly because:

    A)The ball pushes the student backward as the reaction to the student pushing the ball forward.B)Both interaction forces act on the student and cancel completely.C)Friction disappears the instant the ball leaves the hands, so only weight matters.D)The skateboard’s mass becomes negligible during the throw, so $F=ma$ no longer applies.

22. A cart is pulled by two students at angles so their force vectors are not collinear. To find net force on the cart you should:

    A)Add the magnitudes of the two pulls only, treating direction as unimportant while both act on the cart.B)Add the forces as vectors using components or a vector diagram, then sum to a single resultant on the cart.C)Keep only the larger student’s pull on the cart and treat the smaller angled pull as negligible in the calculation.D)Average the two force magnitudes, then multiply that average by the cart’s mass to estimate the net force.

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