How Forces Affect Motion
Chapter 6 — full, step-by-step solutions for every solved example, In-Text “Pause & Ponder”, and end-of-chapter exercise question, with diagrams and clean maths.
Solved Examples
The worked examples from the chapter, fully explained.
Forces along the same straight line simply add when they point the same way and subtract when they point in opposite ways; the net force then points along the larger force.
- (a) Both forces point right (same direction): $F_{net}=10\text{ N}+6\text{ N}=16\text{ N}$ towards the right.
- (b) 10 N right, 6 N left: $F_{net}=10\text{ N}-6\text{ N}=4\text{ N}$ towards the right.
- (c) 6 N right, 10 N left: $F_{net}=10\text{ N}-6\text{ N}=4\text{ N}$ towards the left.
The applied force acts forward; friction acts backward with the same magnitude. These two forces are equal and opposite, so they balance each other and the net force is zero.
By Newton’s first law, an object in motion with zero net force continues to move with constant velocity. So the box neither speeds up nor slows down.
If no net force acts, acceleration is zero, so there are only two possibilities: the object is at rest, or it moves with constant velocity.
- At rest: position stays the same → position–time graph is a horizontal line; velocity is zero → velocity–time graph lies along the time axis.
- Constant velocity: position increases uniformly → position–time graph is a straight inclined line; velocity is unchanging → velocity–time graph is a horizontal line above the axis.
- Find total mass$m = 10 + 10 + 10 = 30\ \text{kg}$
- Weight (downward gravitational force)$F = mg = 30 \times 9.8 = 294\ \text{N}$ (downwards)
- Apply Newton’s first lawTo keep it steady (net force = 0), she must push up with a force equal to the weight.
(i) Push = 50 N. The applied force exactly equals friction (50 N), so the forces are balanced and the net force is zero. The block stays still.
(ii) Push = 55 N.
- Net force$F_{net} = 55 – 50 = 5\ \text{N}$ (forward)
- Acceleration (Newton’s 2nd law)$a = \dfrac{F}{m} = \dfrac{5}{25} = 0.2\ \text{m s}^{-2}$
- Displacement in 2 sUsing $s = ut + \tfrac{1}{2}at^2$ with $u=0$:
$s = 0 + \tfrac{1}{2}\times 0.2 \times (2)^2 = 0.4\ \text{m}$
(i) 0–5 s — line slopes up, so constant acceleration with $u=0$, $v=10\ \text{m s}^{-1}$, $t=5\ \text{s}$:
- Acceleration$v = u + at \Rightarrow 10 = 0 + a(5) \Rightarrow a = 2\ \text{m s}^{-2}$
- Force$F = ma = 1500 \times 2 = 3000\ \text{N}$ towards the east
(ii) 5–10 s — line is parallel to the time axis ⇒ constant velocity ⇒ $a=0$ ⇒ F = 0 (no force).
(iii) 10–15 s — line slopes down, with $u=10$, $v=0$, $t=5\ \text{s}$:
- Acceleration$0 = 10 + a(5) \Rightarrow a = -2\ \text{m s}^{-2}$
- Force$F = 1500 \times (-2) = -3000\ \text{N}$
The forces on the two are equal in magnitude. But the resulting acceleration depends on mass, since $a = \dfrac{F}{m}$.
The Earth’s mass is enormous compared with the fruit, so for the same force its acceleration is extremely tiny — far too small to notice. The fruit, being very light, gets a large acceleration and visibly falls.
By Newton’s third law, the recoil force on the gun is also 2 N.
- Acceleration of the gun$a_{gun} = \dfrac{F}{m_{gun}} = \dfrac{2}{5} = 0.4\ \text{m s}^{-2}$
- Acceleration of the bullet$a_{bullet} = \dfrac{F}{m_{bullet}} = \dfrac{2}{0.1} = 20\ \text{m s}^{-2}$
The forces are equal, but the bullet (smaller mass) accelerates far more than the gun.
In-Text Questions
“Pause and Ponder” questions appearing within the chapter.
Two forces act on the barbell:
- The gravitational force (weight) of the barbell, acting downward.
- The upward force applied by the weightlifter through her hands.
When the barbell is held steady, it is at rest (zero acceleration), so the net force on it must be zero. Hence the two forces are equal in magnitude and opposite in direction.
If the forces were balanced, the clasped hands would stay perfectly still. Because the arms are tilting (moving) towards the front, the hands are accelerating — so a net force acts and the forces are not balanced.
The net force, and therefore the motion, is in the direction of the larger push. So the player who is pushing the joined hands in the direction of the tilt (towards the front / out of the page) is exerting the larger force.
Constant velocity means no change in speed or direction, so the acceleration is zero. By Newton’s first (and second) law, zero acceleration ⇒ zero net force.
- Object remains at rest if at rest.
- Object keeps moving with a constant velocity if already moving.
- Object is moving with a constant acceleration.
No net force means zero acceleration. So an object stays at rest, or keeps its constant velocity. Any acceleration (including a constant one) requires a non-zero net force — so (iii) is impossible.
Several real forces can cancel out so that their combined (net) effect is zero.
Example: A book lying on a table. Gravity pulls it down with its weight, while the table pushes up on it with an equal normal force. These balance, so the net force is zero and the book stays at rest.
Another example: Pushing a box across the floor at a steady speed — the forward push exactly balances the backward friction, giving zero net force and constant velocity.
Constant velocity ⇒ zero acceleration. By $F = ma$, with $a = 0$:
$F = m \times 0 = 0\ \text{N}$
(The mass value is not even needed once we know the velocity is constant.)
From Newton’s second law, $F = ma$. For the same acceleration $a$, the required force is directly proportional to the mass.
So the heavier child needs a larger force, because more force is needed to give a larger mass the same acceleration.
During a jolt or collision, the glass has to lose its momentum. Bubble wrap or hay increases the time over which the glass comes to rest.
A longer stopping time means a smaller deceleration. Since force $= \dfrac{\text{change in momentum}}{\text{time}}$ (i.e. $F = ma$), a smaller acceleration means a smaller force acts on the glass — reducing the chance of breaking.
The pipe pushes a large stream of water forward (out of the nozzle). By Newton’s third law, the water pushes back on the pipe with an equal and opposite force.
This strong backward reaction (recoil/thrust) tends to push the pipe — and the fireperson holding it — backward, so they must brace hard to keep it steady.
It can fire its engine to eject gas (exhaust) in one direction. By Newton’s third law, the ejected gas pushes the spacecraft with an equal and opposite force in the other direction.
This thrust changes the spacecraft’s velocity — speeding it up, slowing it down, or turning it, depending on the direction the gas is expelled.
Exercise
Revise, Reflect, Refine — end-of-chapter questions.
Constant velocity ⇒ zero acceleration ⇒ net force is zero. The only two horizontal forces are the applied force $F$ (forward) and friction $f$ (backward). For them to cancel:
$F – f = 0 \Rightarrow f = F$
- If no net force is applied, the velocity will remain the same / increase / decrease.
- If a net force acts in the direction of motion, the speed will remain the same / increase / decrease.
- If a net force acts opposite to the motion, the speed will remain the same / increase / decrease.
- (i) No net force ⇒ no acceleration ⇒ velocity will remain the same (Newton’s first law).
- (ii) Force along the motion adds to the speed ⇒ speed will increase.
- (iii) Force opposite to the motion slows it down ⇒ speed will decrease.
- P experiences a net force and Q does not.
- P does not experience a net force and Q does.
- Both P and Q experience a net force.
- Neither P nor Q experiences a net force.
Block P: two opposite forces ⇒ net force $= 5 – 4 = 1\ \text{N} \neq 0$ ⇒ P has a net force.
Block Q: moving at constant velocity ⇒ zero acceleration ⇒ net force is zero.
- Forward force (95 oarsmen)$95 \times 200 = 19000\ \text{N}$
- Backward force (5 oarsmen)$5 \times 200 = 1000\ \text{N}$
- Net force$19000 – 1000 = 18000\ \text{N}$ (forward)
- opposite to the force, with acceleration proportional to the force.
- opposite to the force, with acceleration proportional to the mass.
- in the direction of force, with acceleration inversely proportional to the force.
- in the direction of force, with acceleration proportional to the force.
Newton’s second law: $a = \dfrac{F}{m}$. The acceleration is in the same direction as the net force and is directly proportional to it (for fixed mass).
A net force exists only when there is acceleration, i.e. when the velocity (the slope of a position–time graph) is changing — a curved graph.
- A: straight inclined line ⇒ constant velocity ⇒ no net force.
- B: horizontal line ⇒ at rest ⇒ no net force.
- C: curved (slope keeps increasing) ⇒ accelerating ⇒ net force acts.
- D: straight declining line ⇒ constant (negative) velocity ⇒ no net force.
To jump forward, the sailor’s feet push the boat backward. By Newton’s third law, the boat pushes the sailor forward with an equal and opposite force (letting them reach the shore).
The reaction on the boat therefore acts backward, so the boat moves backward — away from the shore, in the direction opposite to the sailor’s jump.
When the athlete lands, they must be brought to rest. A soft mat or sand bed increases the time taken to stop, compared with a hard floor.
A longer stopping time gives a smaller deceleration, and by $F = ma$, a smaller force acts on the athlete’s body. This reduces the risk of injury.
- the loaded cart exerts a larger force on the empty cart.
- the empty cart exerts a larger force on the loaded cart.
- neither cart exerts a force on the other.
- both carts exert equal-magnitude forces on each other.
By Newton’s third law, whenever two objects interact, they exert equal and opposite forces on each other — regardless of their masses or loads. (The loaded cart will accelerate less because of its larger mass, but the forces are equal.)
Read values from the curve: at $m=1\,$kg, $a=10$; at $m=2\,$kg, $a=5$; at $m=4\,$kg, $a=2.5$. In every case:
$F = ma = 1\times10 = 2\times5 = 4\times2.5 = 10\ \text{N}$
The force is the same (10 N) for all masses. So the force–mass graph is a horizontal straight line at $F = 10\ \text{N}$.
- Read the graphAt $t=0$, $v=10\ \text{m s}^{-1}$; at $t=8\ \text{s}$, $v=30\ \text{m s}^{-1}$.
- Acceleration = slope$a = \dfrac{30-10}{8-0} = \dfrac{20}{8} = 2.5\ \text{m s}^{-2}$
- Force$F = ma = 10 \times 2.5 = 25\ \text{N}$
- Convert units$m = 50\ \text{g} = 0.05\ \text{kg}$, $s = 50\ \text{cm} = 0.5\ \text{m}$, $u = 100\ \text{m s}^{-1}$, $v = 0$
- Find accelerationUsing $v^2 = u^2 + 2as$:
$0 = (100)^2 + 2a(0.5) \Rightarrow 0 = 10000 + a \Rightarrow a = -10000\ \text{m s}^{-2}$ - Find force$F = ma = 0.05 \times (-10000) = -500\ \text{N}$
The minus sign shows the force opposes the bullet’s motion.
- Convert speed$v = 108\ \text{km h}^{-1} = 108 \times \dfrac{1000}{3600} = 30\ \text{m s}^{-1}$, $u = 0$
- Acceleration$a = \dfrac{F}{m} = \dfrac{800}{0.4} = 2000\ \text{m s}^{-2}$
- Time of contactUsing $v = u + at$: $30 = 0 + 2000\,t \Rightarrow t = \dfrac{30}{2000} = 0.015\ \text{s}$
- Total opposing force$F = 7 + 3 = 10\ \text{N}$ (backward)
- Acceleration (deceleration)$a = \dfrac{F}{m} = \dfrac{-10}{2} = -5\ \text{m s}^{-2}$
- Distance to stopUsing $v^2 = u^2 + 2as$ with $v=0,\ u=10$:
$0 = (10)^2 + 2(-5)s \Rightarrow 0 = 100 – 10s \Rightarrow s = 10\ \text{m}$
- Express each mass$F = m_1 a_1 \Rightarrow m_1 = \dfrac{F}{a_1}$, $F = m_2 a_2 \Rightarrow m_2 = \dfrac{F}{a_2}$
- Combined systemTotal mass $= m_1 + m_2$. New acceleration:
$a = \dfrac{F}{m_1 + m_2} = \dfrac{F}{\frac{F}{a_1} + \frac{F}{a_2}}$ - Simplify$a = \dfrac{F}{F\left(\frac{1}{a_1} + \frac{1}{a_2}\right)} = \dfrac{1}{\frac{1}{a_1}+\frac{1}{a_2}} = \dfrac{a_1 a_2}{a_1 + a_2}$
The magnetic forces on the two are indeed equal in magnitude. But the resulting acceleration depends on mass, since $a = \dfrac{F}{m}$.
The compass needle is very light, so the same force produces a large acceleration and it swings noticeably. The bar magnet is much heavier, so the equal force gives it only a negligible acceleration — it barely moves.
That’s the full Chapter 6 — solved!
Examples · In-Text “Pause & Ponder” · Exercise “Revise, Reflect, Refine”