Friction & Common Forces — Practice · Grade 12 Science
Chapter: Forces, Rotation & EquilibriumWhat this practice covers
By the end of Friction & Common Forces you should be able to:
- I can distinguish static from kinetic friction and calculate a friction force using f = μN.
- I can identify the normal force, tension, and weight acting on an object.
- I can resolve weight into components on an inclined plane and predict the resulting motion.
10 questions, numeric, multiple choice and written answer, each with a worked explanation.
Vocabulary used: static friction, kinetic friction, coefficient of friction (μ), normal force (N), tension.
Problem 1
Answer and explanation
20 N
f = μₖN = 0.40 × 50 = 20 N.
Problem 2
Answer and explanation
36.75 N
N = mg = 15 × 9.8 = 147 N. f = μₖN = 0.25 × 147 = 36.75 N.
Problem 3
- The maximum static friction is greater than the kinetic friction
- The box loses mass once it moves
- Friction disappears the instant the box moves
- Kinetic friction is greater than static friction
Answer and explanation
The maximum static friction is greater than the kinetic friction
Maximum static friction (μₛN) exceeds kinetic friction (μₖN), so it takes a bigger force to break the box free than to keep it sliding.
Problem 4
- stays exactly the same
- decreases
- immediately becomes zero
- increases
Answer and explanation
decreases
The normal force equals mg cosθ. As θ grows, cosθ shrinks, so the normal force decreases (reaching zero only at a vertical 90° 'ramp').
Problem 5
Answer and explanation
Static friction acts between surfaces that aren't sliding yet and adjusts up to a maximum to prevent motion; kinetic friction acts between surfaces already sliding and is a bit weaker. You feel the switch when pushing a heavy couch: it resists until you push past the static maximum, then suddenly 'gives' and slides more easily under the smaller kinetic friction.
Problem 6
Answer and explanation
3.2 m/s²
Along the slope: a = g(sinθ − μₖcosθ) = 9.8(0.5 − 0.20 × 0.866) = 9.8(0.5 − 0.173) = 9.8 × 0.327 ≈ 3.2 m/s². (Mass cancels out.)
Problem 7
Hints
Static friction instead adjusts up to its limiting value.
Answer and explanation
12 N, enough to balance the required tangential force without exceeding the maximum.
Problem 8
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
It changes the normal force and potentially the sliding tendency; friction depends on N and the motion regime.
Problem 9
Answer and explanation
Recalculate N using the vertical force balance; then update the kinetic-friction estimate and horizontal net force.
Kinetic friction acts during relative sliding. The school approximation fk≈μkN can model a limited range of surface conditions, although real friction may depend on speed, temperature, deformation and wear. On a level table with no vertical acceleration or extra vertical forces, N=mg here. That equality came from the force account, not from the friction law. State the sliding direction before assigning the sign of fk in Newton’s second law.
Recalculate N using the vertical force balance; then update the kinetic-friction estimate and horizontal net force.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 10
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.