Forces, Rotation & Equilibrium — Chapter quiz · Physics & Environmental Science, Grade 12
Use force diagrams, friction, circular motion, springs, and torque to explain motion and balance.
Problem 1
Answer and explanation
6 m/s²
a = ΣF ÷ m = 18 ÷ 3 = 6 m/s².
Problem 2
- the force of gravity
- air resistance
- Newton's third law
- their inertia (Newton's first law)
Answer and explanation
their inertia (Newton's first law)
The passengers were moving with the bus, and their inertia keeps them moving forward when the bus stops — Newton's first law.
Problem 3
- Earth pulling you down and you pulling Earth up
- Your mass and your weight
- Your weight and the normal force from the ground
- The ground pushing up on you and gravity pulling you down
Answer and explanation
Earth pulling you down and you pulling Earth up
A third-law pair acts on two different objects: Earth pulls you down, and you pull Earth up. Weight and normal force act on the same object, so they are not a pair.
Problem 4
Answer and explanation
45 N
Net force = 70 − 25 = 45 N to the right.
Problem 5
Answer and explanation
2500 N
First find acceleration: a = (25 − 15) ÷ 4 = 2.5 m/s². Then ΣF = ma = 1,000 × 2.5 = 2,500 N.
Problem 6
Answer and explanation
14 N
f = μₖN = 0.35 × 40 = 14 N.
Problem 7
Answer and explanation
4.9 m/s²
On a frictionless ramp, a = g sinθ = 9.8 × sin 30° = 9.8 × 0.5 = 4.9 m/s². (Mass does not matter.)
Problem 8
Answer and explanation
9 m/s²
a = v²/r = 15² ÷ 25 = 225 ÷ 25 = 9 m/s².
Problem 9
- backward, opposing the motion
- toward the center of the circle
- along the direction of motion
- away from the center, outward
Answer and explanation
toward the center of the circle
Centripetal acceleration always points toward the center of the circular path.
Problem 10
Answer and explanation
3.6 N
F = mv²/r = 0.2 × 3² ÷ 0.5 = 0.2 × 9 ÷ 0.5 = 1.8 ÷ 0.5 = 3.6 N.
Problem 11
- 3 times as strong
- 9 times as strong
- 1/9 as strong
- 1/3 as strong
Answer and explanation
1/9 as strong
Gravitation is inverse-square (F ∝ 1/r²). Tripling r gives a force 1/3² = 1/9 as strong.
Problem 12
Answer and explanation
The astronaut and the station are both in continuous free-fall around Earth — gravity is pulling on both equally, so they fall together at the same rate. With nothing pushing up on the astronaut (no normal force), they feel weightless, even though gravity is still supplying the centripetal force that keeps them in orbit.
Problem 13
Answer and explanation
a=Fnet/m=2 m/s² along the net force. Action/reaction forces act on different objects, so they do not cancel in one object’s diagram.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 14
Answer and explanation
Net=12 N east; a=3 m/s² east. Its third-law partner is the cart’s 18 N force on the pulling agent, not the 6 N resistance on the cart.
Net=12 N east; a=3 m/s² east.
Its third-law partner is the cart’s 18 N force on the pulling agent, not the 6 N resistance on the cart.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 15
Answer and explanation
Net 12N right; a=12/4=3 m/s² right. Draw only forces exerted on the selected object; motion arrows are not forces.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 16
Answer and explanation
N=29.4−7.5=21.9 N. Horizontal net=15cos30°−4≈8.99 N; ax≈3.00 m/s². The tension is counted once through its components.
N=29.4−7.5=21.9 N.
Horizontal net=15cos30°−4≈8.99 N; ax≈3.00 m/s².
The tension is counted once through its components.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 17
Answer and explanation
Static friction is 2N opposite the push, so net force is zero. It need not reach 5N. Static friction adjusts up toμsN; it is not always equal to that maximum.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 18
Answer and explanation
N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction. During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.
N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction.
During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 19
Answer and explanation
ac=v²/r=8 m/s². Speed can be constant while velocity direction changes. Centripetal force is the net inward force, not an extra interaction added to gravity or tension.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 20
Answer and explanation
ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N. Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.
ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N.
Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 21
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
0.4 J
.5×80×.01=.4.
Problem 22
Answer and explanation
Zero resultant force and zero resultant torque.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 23
Answer and explanation
Torque balance: F×2=60×0.50, so F=15 N upward. Vertical balance gives pivot reaction=60−15=45 N upward. Both balances are necessary.
Torque balance: F×2=60×0.50, so F=15 N upward.
Vertical balance gives pivot reaction=60−15=45 N upward.
Both balances are necessary.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.