01 · Net force first
A 4 kg particle has forces 28 N right and 12 N left. Find its acceleration.
Hint
Resolve rightward before applying the law.
Worked solution
28 − 12 = 4a, so a = 4 m/s² right.
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Newton’s second law, mass and weight
Use resultant force equals mass times acceleration, convert units, distinguish mass from weight and solve signed force equations.
Before you startForce diagrams, signed resultants and SI units.
01 / Apply Newton’s second law to the resultant
ΣF = ma connects force and acceleration.
For constant mass, resultant force = mass × acceleration.
Use kilograms for mass, metres per second squared for acceleration and newtons for force. Along a chosen axis, use signed components. Since mass is positive, the resultant and acceleration have the same direction. This does not determine the current velocity.
Resultant divided by massExplore
Take right as positive. The force shown is already the total horizontal resultant. Compare cases with the same resultant and different masses, then reverse or remove the resultant. The acceleration arrow is separate from the force diagram.
02 / Add the forces before dividing by mass
An individual driving force is not necessarily the resultant.
A 3 kg particle has a 24 N rightward pull and 9 N leftward resistance.Worked example
ΣF = 24 − 9 = 15 N
Take right as positive.
15 = 3a, so a = 5 m/s² right
Use the net force, not 24 N alone.
Vertical forces may balance independently
A zero vertical acceleration does not prevent horizontal acceleration.
Watch: the same resultant accelerates smaller masses more
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02 · Force from acceleration
A 2.5 kg particle has acceleration −1.2 m/s² along the chosen positive axis. Find the resultant component.
Hint
Multiply signed acceleration by mass.
Worked solution
ΣF = 2.5(−1.2) = −3 N, meaning 3 N opposite the positive direction.
03 / Convert mass into kilograms before using newtons
A gram value changes the numerical calculation by a factor of 1000.
03 · Gram conversion
A 350 g particle experiences resultant force 1.4 N. Find its acceleration magnitude.
Hint
350 g = 0.350 kg.
Worked solution
a = 1.4/0.350 = 4 m/s².
04 · Unit definition
What resultant force gives a 1 kg mass an acceleration of 1 m/s²?
Hint
Use ΣF = ma.
Worked solution
1 N. In SI units, 1 N = 1 kg m/s².
05 · A unit error
A learner divides 1.4 by 350 and reports 0.004 m/s² for question 3. Explain the error.
Hint
The force is in newtons but the mass was entered in grams.
Worked solution
The numerical mass must be 0.350 kg, not 350. The reported acceleration is 1000 times too small.
04 / Mass and weight are different quantities
Weight is a gravitational force: W = mg.
Mass is measured in kg; weight is measured in N.
With g = 9.8 m/s², the magnitude of weight is 9.8m N for mass m kg. A body keeps the same mass when the local gravitational field changes, but its weight changes. A supported body can have nonzero weight while its total resultant is zero.
06 · Weight on Earth
Find the weight of a 0.6 kg particle using g = 9.8 m/s².
Hint
Multiply mass by g.
Worked solution
W = 0.6 × 9.8 = 5.88 N downward.
07 · Different gravity
A 6 kg object is where g = 1.62 m/s². Find its mass and weight there.
Hint
Only the gravitational force changes.
Worked solution
Mass remains 6 kg. Weight = 6 × 1.62 = 9.72 N.
08 · Mass from weight
An object weighs 78.4 N where g = 9.8 m/s². Find its mass.
Hint
Rearrange W = mg.
Worked solution
m = 78.4/9.8 = 8 kg.
05 / Keep a single positive direction throughout the equation
A negative answer describes direction, not negative mass.
09 · Vertical resultant
A 2 kg particle experiences 18 N upward tension and its weight. Take upward positive and g = 9.8. Find its acceleration.
Hint
Weight is 19.6 N downward.
Worked solution
18 − 19.6 = 2a, so a = −0.8 m/s²: 0.8 m/s² downward. Its current velocity has not been specified.
10 · Vector form
A 2 kg particle has resultant 6i − 8j N. Find its acceleration vector and magnitude.
Hint
Divide each force component by the same mass.
Worked solution
a = 3i − 4j m/s²; magnitude √(9 + 16) = 5 m/s².
06 / Use measured acceleration to infer a force or mass
The unknown may be inside the resultant.
11 · Unknown resistance
A 6 kg particle is pulled right by 30 N and accelerates right at 2 m/s². Find leftward resistance R.
Hint
Write 30 − R = 6 × 2.
Worked solution
R = 18 N. The resultant is 12 N; resistance is not the same quantity.
12 · Unknown mass
A resultant of magnitude 14 N gives acceleration magnitude 3.5 m/s². Find the mass.
Hint
Rearrange m = |ΣF|/|a|.
Worked solution
m = 14/3.5 = 4 kg.
13 · No mass information
A particle has zero resultant and zero acceleration. Can 0 = m × 0 determine its mass?
Hint
Does any positive m violate this equation?
Worked solution
No. Every positive mass satisfies it; further information is required. Dividing zero by zero does not give a mass.
07 / State when the acceleration is constant
SUVAT needs a constant resultant as well as constant mass.
14 · Changing resistance
A particle has constant driving force and constant mass, but resistance grows as its speed grows. Must its acceleration be constant?
Hint
Check the resultant, not just the driving force.
Worked solution
No. Driving force minus resistance changes, so the resultant and acceleration change. A single constant-acceleration equation cannot be assumed for the whole interval.
Draw every relevant external force on the chosen body. A frictionless surface removes friction, not necessarily gravity or the normal reaction. Use ΣF = ma separately along perpendicular directions.
08 / Draw, resolve, then apply the law
Use kilograms and a consistent sign convention.
Calculate the resultant before using ΣF = ma. Keep mass distinct from weight, infer acceleration direction from its sign, and check whether the force model justifies constant acceleration over the interval.
Section 1 of 8 · Apply Newton’s second law to the resultant