Module 2: Dynamics

Explain why objects move using Newton's laws: draw the forces, apply F = ma, handle friction and inclined planes, and follow energy, work, power and momentum through collisions.

Pillar 1 of 3

Forces

Kinematics described how things move; dynamics explains why. It begins with forces, the pushes and pulls that change motion. Learn the common forces, draw them cleanly on free-body diagrams, combine them into a net force, and see how forces always come in pairs.

Force as a Vector

A force is a push or a pull, measured in newtons (N). It is a vector, so it has both a size and a direction. Forces can make an object speed up, slow down, change direction, or change shape.

Contact forces

Act only when objects touch: the normal force, friction, tension, and applied pushes.

Field forces

Act at a distance through a field: gravity, and electric and magnetic forces.

Worked Example

Two forces act on a box along a straight line: $8\text{ N}$ to the right and $\;5\text{ N}$ to the right. Find the single combined force. Then find it if the $\;5\text{ N}$ force instead acts to the left.

Solution (take right as positive):

$\text{Both right: } 8 + 5 = \mathbf{13\text{ N right}}$

$\text{Opposite: } 8 - 5 = \mathbf{3\text{ N right}}$

Because force is a vector, direction decides whether the forces add or partly cancel.

Practice Question

Q: Classify each force as a contact force or a field force: gravity, tension, friction, the magnetic force.

Show Answer

Contact (objects must touch): tension and friction.

Field (act at a distance): gravity and the magnetic force.

Common Forces

A handful of forces appear again and again. The most important is weight, the pull of gravity on a mass.

Weight (W)

Gravity's pull, straight down: $W = mg$ .

Normal force (N)

The support force from a surface, perpendicular to it.

Tension (T)

The pull transmitted along a rope, string or cable.

Friction (f)

Resists sliding, acting along the surface opposite the motion.

Weight

W = mg

$m$ = mass (kg), $g \approx 9.8\text{ m/s}^2$

Worked Example

Find the weight of a $5.0\text{ kg}$ bag. ( $g = 9.8\text{ m/s}^2$ )

Solution:

$W = mg = (5.0)(9.8) = 49\text{ N}$ downward

Practice Question

Q: An astronaut has a mass of $80\text{ kg}$ . Find their weight on Earth ( $g = 9.8\text{ m/s}^2$ ) and on the Moon ( $g = 1.6\text{ m/s}^2$ ). What stays the same?

Show Answer

$W_{\text{Earth}} = 80 \times 9.8 = 784\text{ N}$

$W_{\text{Moon}} = 80 \times 1.6 = 128\text{ N}$

The mass ( $80\text{ kg}$ ) is unchanged; only the weight differs, because $g$ is different.

Free-Body Diagrams

A free-body diagram shows every force acting on one object as an arrow from its centre. It is the single most important habit in dynamics: isolate the object, draw and label each force in the correct direction, and only then do the maths.

A free-body diagram of a box being pushed along the ground: weight and normal force balance vertically, while the applied push competes with friction.

Worked Example

A $2.0\text{ kg}$ book rests in equilibrium on a flat table. List the forces on the book and state the size of each. ( $g = 9.8\text{ m/s}^2$ )

Solution:

Weight (down): $W = mg = (2.0)(9.8) = 19.6\text{ N}$

Normal force (up): the book is in equilibrium, so $N = W = 19.6\text{ N}$

Net force: $\sum F = 0$

Practice Question

Q: A book sits on a table. Is the table's normal force the Newton's third law reaction to the book's weight?

Show Answer

No. Weight's reaction pair is the book pulling the Earth upward. The normal force pairs with the book pushing down on the table.

Weight and normal force are equal here only because the book is in equilibrium, not because they are an action-reaction pair (they act on the same object, so they never could be).

Net Force and Equilibrium

The net force is the vector sum of all forces on an object. If the net force is zero the object is in equilibrium, meaning it stays at rest or keeps moving at constant velocity.

Equilibrium Condition

\sum F = 0

Worked Example

A crate is pushed with $30\text{ N}$ to the right while friction of $\;12\text{ N}$ acts to the left. What is the net horizontal force?

Solution (take right as positive):

$\sum F = 30 - 12 = 18\text{ N to the right}$

Since the net force is not zero, the crate accelerates to the right.

Practice Question

Q: Three horizontal forces act on a trolley: $40\text{ N}$ east, $\;30\text{ N}$ west, and $20\text{ N}$ east. Find the net force. Is the trolley in equilibrium?

Show Answer

Take east as positive:

$\sum F = 40 - 30 + 20 = 30\text{ N east}$

The net force is not zero, so it is not in equilibrium; the trolley accelerates east.

Newton's Third Law

For every action there is an equal and opposite reaction. Forces always come in pairs, and crucially the two forces act on different objects, which is why they do not cancel out.

Action and Reaction

\vec{F}_{A\text{ on }B} = -\vec{F}_{B\text{ on }A}

When two skaters push off each other, the force on A and the force on B are equal in size and opposite in direction, but act on different people.

Worked Example

Skater A (mass $50\text{ kg}$ ) pushes on skater B (mass $75\text{ kg}$ ), and A feels a force of $150\text{ N}$ . What force acts on B, and how do their accelerations compare?

Solution:

By Newton's third law the force on B is equal and opposite: $\mathbf{150\text{ N}}$ .

$a_A = \frac{F}{m_A} = \frac{150}{50} = 3.0\text{ m/s}^2$

$a_B = \frac{F}{m_B} = \frac{150}{75} = 2.0\text{ m/s}^2$

Equal forces, but the lighter skater accelerates more.

Practice Question

Q: When you walk forward, what is the action-reaction pair, and which force actually pushes you along?

Show Answer

Your foot pushes backward on the ground (action); the ground pushes forward on your foot (reaction).

It is the reaction force from the ground, acting on you, that drives you forward.

Pillar 2 of 3

Forces, Acceleration and Energy

Newton's laws tie force to motion. Here you use $F = ma$ to predict acceleration, handle friction and inclined planes, and follow the energy: the work a force does, the kinetic energy it produces, and the power at which it is delivered.

Newton's First Law (Inertia)

An object stays at rest, or keeps moving at constant velocity, unless a net force acts on it. This resistance to any change in motion is called inertia, and it increases with mass. A stationary object does not spontaneously move, and a moving object does not slow down or turn on its own; a force is always responsible.

At rest

A stationary object stays put until a net force starts it moving.

In motion

A moving object keeps the same speed and direction until a net force changes it.

Worked Example

A $1200\text{ kg}$ car travels in a straight line at a constant $20\text{ m/s}$ . What is the net force on it?

Solution:

Constant velocity means $a = 0$ , so $\sum F = ma = 1200 \times 0 = \mathbf{0\text{ N}}$ .

The driving force exactly balances friction and air resistance, so the forces cancel.

Practice Question

Q: Why do passengers lurch forward when a bus brakes suddenly?

Show Answer

By inertia, their bodies tend to keep moving forward at the original speed while the bus decelerates beneath them. They keep going until a force (a seatbelt, or friction with the seat) slows them too.

Newton's Second Law

When there is a net force, it produces an acceleration in the same direction, proportional to the force and inversely proportional to the mass.

Newton's Second Law

\sum F = ma

$\sum F$ = net force (N), $m$ = mass (kg), $a$ = acceleration (m/s²)

Worked Example

A $1200\text{ kg}$ car experiences a net driving force of $3600\text{ N}$ . Find its acceleration.

Solution:

$a = \frac{\sum F}{m} = \frac{3600}{1200} = 3.0\text{ m/s}^2$

Practice Question

Q: A $2.0\text{ kg}$ trolley is pushed with $10\text{ N}$ while friction of $4\text{ N}$ opposes it. Find its acceleration.

Show Answer

Find the net force first, then apply $\sum F = ma$ :

$\sum F = 10 - 4 = 6\text{ N}$

$a = \frac{\sum F}{m} = \frac{6}{2.0} = 3.0\text{ m/s}^2$

Friction

Friction opposes sliding between surfaces. Its size grows with how hard the surfaces are pressed together, that is, with the normal force. Static friction holds a stationary object in place up to a maximum; kinetic friction acts once it slides.

Friction Force

f = \mu N

$\mu$ = coefficient of friction, $N$ = normal force (N)

Worked Example

A $10\text{ kg}$ box sits on level ground with a coefficient of friction $\;\mu = 0.30$ . What friction force must be overcome to slide it?

Solution (on level ground $N = mg$ ):

$N = mg = (10)(9.8) = 98\text{ N}$

$f = \mu N = (0.30)(98) = 29.4\text{ N}$

Practice Question

Q: A $20\text{ kg}$ crate on level ground has $\mu = 0.25$ and is pushed with $60\text{ N}$ . Find the kinetic friction, the net force and the acceleration. ( $g = 9.8\text{ m/s}^2$ )

Show Answer

$f = \mu mg = (0.25)(20)(9.8) = 49\text{ N}$

$\sum F = 60 - 49 = 11\text{ N}$

$a = \frac{11}{20} = 0.55\text{ m/s}^2$

Inclined Planes

On a slope, the trick is to resolve the weight into components along and perpendicular to the surface. For a slope at angle $\theta$ , the pull down the slope is $mg\sin\theta$ and the part pressing into the surface is $\;mg\cos\theta$ (which the normal force balances).

Weight Components on a Slope

F_\parallel = mg\sin\theta

F_\perp = mg\cos\theta

Worked Example

A $5.0\text{ kg}$ block is released on a frictionless $30^\circ$ incline. Find the force pulling it down the slope and its acceleration.

The weight mg resolves into mg sin θ down the slope and mg cos θ into the surface.

Solution:

$F_\parallel = mg\sin\theta = (5.0)(9.8)\sin 30^\circ = 24.5\text{ N}$

$a = g\sin\theta = 9.8\sin 30^\circ = 4.9\text{ m/s}^2$

The mass cancels: every object slides down a frictionless slope at the same rate.

Practice Question

Q: A $3.0\text{ kg}$ block sits on a frictionless $25^\circ$ slope. Find (a) the component of weight down the slope and (b) its acceleration. ( $g = 9.8\text{ m/s}^2$ )

Show Answer

$F_\parallel = mg\sin\theta = (3.0)(9.8)\sin 25^\circ = 12.4\text{ N}$

$a = g\sin\theta = 9.8\sin 25^\circ = 4.1\text{ m/s}^2$

Work

Work is done when a force moves an object, and it transfers energy. Only the part of the force along the motion does work, so an angle $\theta$ between force and displacement brings in a $\cos\theta$ factor. Work is measured in joules (J).

Work Done

W = Fs\cos\theta

Worked Example

A box is pulled $10\text{ m}$ across the floor by a $50\text{ N}$ force directed at $30^\circ$ above the horizontal. How much work is done?

Only the horizontal part of the force, F cos θ, does work over the displacement s.

Solution:

$W = Fs\cos\theta = (50)(10)\cos 30^\circ = 433\text{ J}$

Practice Question

Q: A $20\text{ N}$ force pushes a box $5\text{ m}$ in the direction of the force. How much work is done? What if the same force acted at $90^\circ$ to the motion?

Show Answer

$W = Fs\cos 0^\circ = (20)(5)(1) = 100\text{ J}$

At $90^\circ$ , $\cos 90^\circ = 0$ , so $W = 0\text{ J}$ . A force perpendicular to the motion does no work.

Kinetic Energy and the Work-Energy Theorem

A moving object carries kinetic energy. The work-energy theorem connects the two ideas: the net work done on an object equals the change in its kinetic energy.

Kinetic Energy and Work-Energy Theorem

KE = \tfrac{1}{2}mv^2

W_{\text{net}} = \Delta KE

Worked Example

How much work is needed to accelerate a $1000\text{ kg}$ car from rest to $\;20\text{ m/s}$ ?

Solution (work = gain in kinetic energy):

$W = \Delta KE = \tfrac{1}{2}mv^2 - 0 = \tfrac{1}{2}(1000)(20)^2$

$W = 2.0\times10^5\text{ J} = 200\text{ kJ}$

Practice Question

Q: A $0.20\text{ kg}$ ball moves at $15\text{ m/s}$ . Find its kinetic energy. What happens to the kinetic energy if its speed doubles?

Show Answer

$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.20)(15)^2 = 22.5\text{ J}$

Because $KE \propto v^2$ , doubling the speed quadruples the kinetic energy to $90\text{ J}$ .

Power

Power is how fast work is done, or energy is transferred, measured in watts (W). For a steady force moving an object at speed $v$ , power can also be written as $\;P = Fv$ .

Power

P = \frac{W}{t} = Fv

Worked Example

A motor does $60{,}000\text{ J}$ of work in $4.0\text{ s}$ . What is its power output?

Solution:

$P = \frac{W}{t} = \frac{60{,}000}{4.0} = 15{,}000\text{ W} = 15\text{ kW}$

Practice Question

Q: A car engine provides a steady driving force of $500\text{ N}$ while the car cruises at $30\text{ m/s}$ . What power is the engine delivering?

Show Answer

$P = Fv = (500)(30) = 15{,}000\text{ W} = 15\text{ kW}$

Pillar 3 of 3

Momentum, Energy and Simple Systems

When objects interact, momentum is the quantity that is passed between them. Learn momentum and impulse, then use the conservation of momentum to analyse collisions, and tell elastic collisions apart from inelastic ones.

Momentum

Momentum measures how hard it is to stop a moving object. It is a vector, the product of mass and velocity, measured in $\text{kg·m/s}$ .

Momentum

p = mv

Worked Example

Find the momentum of a $1500\text{ kg}$ car travelling at $20\text{ m/s}$ .

Solution:

$p = mv = (1500)(20) = 3.0\times10^4\text{ kg·m/s}$

Practice Question

Q: A $0.15\text{ kg}$ cricket ball travels at $40\text{ m/s}$ . Find its momentum. A $60\text{ kg}$ cyclist has the same momentum, how fast are they moving?

Show Answer

$p = mv = (0.15)(40) = 6\text{ kg·m/s}$

$v = \frac{p}{m} = \frac{6}{60} = 0.1\text{ m/s}$

A light, fast ball can carry the same momentum as a heavy, slow rider.

Impulse

Impulse is the change in momentum a force produces, equal to the force multiplied by the time it acts. On a force-time graph, the impulse is simply the area under the graph. This is why airbags and crumple zones save lives: extending the time of a collision lowers the force for the same change in momentum.

Impulse

J = Ft = \Delta p = m\Delta v

Worked Example

A constant force of $200\text{ N}$ acts on a ball for $0.50\text{ s}$ . Find the impulse delivered, and hence the change in the ball's momentum.

The impulse is the area under the force-time graph: 200 N × 0.5 s = 100 N·s.

Solution:

$J = Ft = (200)(0.50) = 100\text{ N·s}$

$\Delta p = J = 100\text{ kg·m/s}$

Practice Question

Q: A $0.058\text{ kg}$ tennis ball is served from rest to $30\text{ m/s}$ . Find the impulse. If the racket is in contact for $0.005\text{ s}$ , find the average force.

Show Answer

$J = m\Delta v = (0.058)(30) = 1.74\text{ N·s}$

$F = \frac{J}{t} = \frac{1.74}{0.005} = 348\text{ N}$

A longer contact time would lower this force, the same idea behind crumple zones.

Conservation of Momentum

In an isolated system (no external forces), the total momentum before an interaction equals the total momentum after it. This single principle solves almost every collision and explosion problem.

Conservation of Momentum

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Worked Example

A $2.0\text{ kg}$ trolley moving at $3.0\text{ m/s}$ collides with a stationary $1.0\text{ kg}$ trolley and they lock together. Find their common velocity afterward.

Momentum is conserved: 6 + 0 kg·m/s before equals 6 kg·m/s after, as the 2 kg trolley locks onto the stationary 1 kg trolley to move off at 2 m/s.

Solution (they stick, so share one final velocity):

$m_1u_1 + m_2u_2 = (m_1 + m_2)v$

$(2.0)(3.0) + 0 = (3.0)v$

$v = \frac{6.0}{3.0} = 2.0\text{ m/s}$

Practice Question

Q: A $60\text{ kg}$ astronaut floats at rest and throws a $2.0\text{ kg}$ tool at $5.0\text{ m/s}$ . Find the astronaut's recoil velocity.

Show Answer

Total momentum starts at zero and is conserved:

$0 = (2.0)(5.0) + (60)v$

$v = -\frac{10}{60} = -0.17\text{ m/s}$

The astronaut drifts at $0.17\text{ m/s}$ in the opposite direction to the tool.

Elastic and Inelastic Collisions

Momentum is conserved in every collision. Kinetic energy is not: whether it is conserved separates the two types.

Elastic

Both momentum and kinetic energy are conserved. Objects bounce apart cleanly (idealised, like billiard balls).

Inelastic

Momentum is conserved but kinetic energy is not; some is lost to heat, sound and deformation. If the objects stick together it is perfectly inelastic.

Worked Example

In the collision above, a $2.0\text{ kg}$ trolley at $3.0\text{ m/s}$ sticks to a stationary $1.0\text{ kg}$ trolley and they move off together at $\;2.0\text{ m/s}$ . Show that kinetic energy is lost, confirming the collision is inelastic.

Compare kinetic energy before and after:

$KE_{\text{before}} = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\text{ J}$

$KE_{\text{after}} = \tfrac{1}{2}(3.0)(2.0)^2 = 6.0\text{ J}$

$\Delta KE = 6.0 - 9.0 = -3.0\text{ J}$

Momentum was conserved, but $3.0\text{ J}$ of kinetic energy became heat, sound and deformation, so the collision is inelastic.

Practice Question

Q: Two lumps of clay collide head-on and stick together. Is momentum conserved? Is kinetic energy conserved?

Show Answer

Momentum: yes, it is always conserved when no external forces act.

Kinetic energy: no. This is a perfectly inelastic collision, so some kinetic energy becomes heat and permanent deformation of the clay.

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Pillar 1 of 3

Forces

Force as a Vector

A force is a push or a pull, measured in newtons (N). It is a vector, so it has both a size and a direction. Forces can make an object speed up, slow down, change direction, or change shape.

Contact forces

Act only when objects touch: the normal force, friction, tension, and applied pushes.

Field forces

Act at a distance through a field: gravity, and electric and magnetic forces.

Worked Example

Solution (take right as positive):

$\text{Both right: } 8 + 5 = \mathbf{13\text{ N right}}$

$\text{Opposite: } 8 - 5 = \mathbf{3\text{ N right}}$

Because force is a vector, direction decides whether the forces add or partly cancel.

Practice Question

Q: Classify each force as a contact force or a field force: gravity, tension, friction, the magnetic force.

Show Answer

Contact (objects must touch): tension and friction.

Field (act at a distance): gravity and the magnetic force.

Common Forces

A handful of forces appear again and again. The most important is weight, the pull of gravity on a mass.

Weight (W)

Gravity's pull, straight down: $W = mg$ .

Normal force (N)

The support force from a surface, perpendicular to it.

Tension (T)

The pull transmitted along a rope, string or cable.

Friction (f)

Resists sliding, acting along the surface opposite the motion.

Weight

W = mg

$m$ = mass (kg), $g \approx 9.8\text{ m/s}^2$

Worked Example

Find the weight of a $5.0\text{ kg}$ bag. ( $g = 9.8\text{ m/s}^2$ )

Solution:

$W = mg = (5.0)(9.8) = 49\text{ N}$ downward

Practice Question

Q: An astronaut has a mass of $80\text{ kg}$ . Find their weight on Earth ( $g = 9.8\text{ m/s}^2$ ) and on the Moon ( $g = 1.6\text{ m/s}^2$ ). What stays the same?

Show Answer

$W_{\text{Earth}} = 80 \times 9.8 = 784\text{ N}$

$W_{\text{Moon}} = 80 \times 1.6 = 128\text{ N}$

The mass ( $80\text{ kg}$ ) is unchanged; only the weight differs, because $g$ is different.

Free-Body Diagrams

A free-body diagram of a box being pushed along the ground: weight and normal force balance vertically, while the applied push competes with friction.

Worked Example

A $2.0\text{ kg}$ book rests in equilibrium on a flat table. List the forces on the book and state the size of each. ( $g = 9.8\text{ m/s}^2$ )

Solution:

Weight (down): $W = mg = (2.0)(9.8) = 19.6\text{ N}$

Normal force (up): the book is in equilibrium, so $N = W = 19.6\text{ N}$

Net force: $\sum F = 0$

Practice Question

Q: A book sits on a table. Is the table's normal force the Newton's third law reaction to the book's weight?

Show Answer

No. Weight's reaction pair is the book pulling the Earth upward. The normal force pairs with the book pushing down on the table.

Weight and normal force are equal here only because the book is in equilibrium, not because they are an action-reaction pair (they act on the same object, so they never could be).

Net Force and Equilibrium

The net force is the vector sum of all forces on an object. If the net force is zero the object is in equilibrium, meaning it stays at rest or keeps moving at constant velocity.

Equilibrium Condition

\sum F = 0

Worked Example

A crate is pushed with $30\text{ N}$ to the right while friction of $\;12\text{ N}$ acts to the left. What is the net horizontal force?

Solution (take right as positive):

$\sum F = 30 - 12 = 18\text{ N to the right}$

Since the net force is not zero, the crate accelerates to the right.

Practice Question

Q: Three horizontal forces act on a trolley: $40\text{ N}$ east, $\;30\text{ N}$ west, and $20\text{ N}$ east. Find the net force. Is the trolley in equilibrium?

Show Answer

Take east as positive:

$\sum F = 40 - 30 + 20 = 30\text{ N east}$

The net force is not zero, so it is not in equilibrium; the trolley accelerates east.

Newton's Third Law

For every action there is an equal and opposite reaction. Forces always come in pairs, and crucially the two forces act on different objects, which is why they do not cancel out.

Action and Reaction

\vec{F}_{A\text{ on }B} = -\vec{F}_{B\text{ on }A}

When two skaters push off each other, the force on A and the force on B are equal in size and opposite in direction, but act on different people.

Worked Example

Skater A (mass $50\text{ kg}$ ) pushes on skater B (mass $75\text{ kg}$ ), and A feels a force of $150\text{ N}$ . What force acts on B, and how do their accelerations compare?

Solution:

By Newton's third law the force on B is equal and opposite: $\mathbf{150\text{ N}}$ .

$a_A = \frac{F}{m_A} = \frac{150}{50} = 3.0\text{ m/s}^2$

$a_B = \frac{F}{m_B} = \frac{150}{75} = 2.0\text{ m/s}^2$

Equal forces, but the lighter skater accelerates more.

Practice Question

Q: When you walk forward, what is the action-reaction pair, and which force actually pushes you along?

Show Answer

Your foot pushes backward on the ground (action); the ground pushes forward on your foot (reaction).

It is the reaction force from the ground, acting on you, that drives you forward.

Pillar 2 of 3

Forces, Acceleration and Energy

Newton's First Law (Inertia)

At rest

A stationary object stays put until a net force starts it moving.

In motion

A moving object keeps the same speed and direction until a net force changes it.

Worked Example

A $1200\text{ kg}$ car travels in a straight line at a constant $20\text{ m/s}$ . What is the net force on it?

Solution:

Constant velocity means $a = 0$ , so $\sum F = ma = 1200 \times 0 = \mathbf{0\text{ N}}$ .

The driving force exactly balances friction and air resistance, so the forces cancel.

Practice Question

Q: Why do passengers lurch forward when a bus brakes suddenly?

Show Answer

Newton's Second Law

When there is a net force, it produces an acceleration in the same direction, proportional to the force and inversely proportional to the mass.

Newton's Second Law

\sum F = ma

$\sum F$ = net force (N), $m$ = mass (kg), $a$ = acceleration (m/s²)

Worked Example

A $1200\text{ kg}$ car experiences a net driving force of $3600\text{ N}$ . Find its acceleration.

Solution:

$a = \frac{\sum F}{m} = \frac{3600}{1200} = 3.0\text{ m/s}^2$

Practice Question

Q: A $2.0\text{ kg}$ trolley is pushed with $10\text{ N}$ while friction of $4\text{ N}$ opposes it. Find its acceleration.

Show Answer

Find the net force first, then apply $\sum F = ma$ :

$\sum F = 10 - 4 = 6\text{ N}$

$a = \frac{\sum F}{m} = \frac{6}{2.0} = 3.0\text{ m/s}^2$

Friction

Friction Force

f = \mu N

$\mu$ = coefficient of friction, $N$ = normal force (N)

Worked Example

A $10\text{ kg}$ box sits on level ground with a coefficient of friction $\;\mu = 0.30$ . What friction force must be overcome to slide it?

Solution (on level ground $N = mg$ ):

$N = mg = (10)(9.8) = 98\text{ N}$

$f = \mu N = (0.30)(98) = 29.4\text{ N}$

Practice Question

Q: A $20\text{ kg}$ crate on level ground has $\mu = 0.25$ and is pushed with $60\text{ N}$ . Find the kinetic friction, the net force and the acceleration. ( $g = 9.8\text{ m/s}^2$ )

Show Answer

$f = \mu mg = (0.25)(20)(9.8) = 49\text{ N}$

$\sum F = 60 - 49 = 11\text{ N}$

$a = \frac{11}{20} = 0.55\text{ m/s}^2$

Inclined Planes

Weight Components on a Slope

F_\parallel = mg\sin\theta

F_\perp = mg\cos\theta

Worked Example

A $5.0\text{ kg}$ block is released on a frictionless $30^\circ$ incline. Find the force pulling it down the slope and its acceleration.

The weight mg resolves into mg sin θ down the slope and mg cos θ into the surface.

Solution:

$F_\parallel = mg\sin\theta = (5.0)(9.8)\sin 30^\circ = 24.5\text{ N}$

$a = g\sin\theta = 9.8\sin 30^\circ = 4.9\text{ m/s}^2$

The mass cancels: every object slides down a frictionless slope at the same rate.

Practice Question

Q: A $3.0\text{ kg}$ block sits on a frictionless $25^\circ$ slope. Find (a) the component of weight down the slope and (b) its acceleration. ( $g = 9.8\text{ m/s}^2$ )

Show Answer

$F_\parallel = mg\sin\theta = (3.0)(9.8)\sin 25^\circ = 12.4\text{ N}$

$a = g\sin\theta = 9.8\sin 25^\circ = 4.1\text{ m/s}^2$

Work

Work Done

W = Fs\cos\theta

Worked Example

A box is pulled $10\text{ m}$ across the floor by a $50\text{ N}$ force directed at $30^\circ$ above the horizontal. How much work is done?

Only the horizontal part of the force, F cos θ, does work over the displacement s.

Solution:

$W = Fs\cos\theta = (50)(10)\cos 30^\circ = 433\text{ J}$

Practice Question

Q: A $20\text{ N}$ force pushes a box $5\text{ m}$ in the direction of the force. How much work is done? What if the same force acted at $90^\circ$ to the motion?

Show Answer

$W = Fs\cos 0^\circ = (20)(5)(1) = 100\text{ J}$

At $90^\circ$ , $\cos 90^\circ = 0$ , so $W = 0\text{ J}$ . A force perpendicular to the motion does no work.

Kinetic Energy and the Work-Energy Theorem

A moving object carries kinetic energy. The work-energy theorem connects the two ideas: the net work done on an object equals the change in its kinetic energy.

Kinetic Energy and Work-Energy Theorem

KE = \tfrac{1}{2}mv^2

W_{\text{net}} = \Delta KE

Worked Example

How much work is needed to accelerate a $1000\text{ kg}$ car from rest to $\;20\text{ m/s}$ ?

Solution (work = gain in kinetic energy):

$W = \Delta KE = \tfrac{1}{2}mv^2 - 0 = \tfrac{1}{2}(1000)(20)^2$

$W = 2.0\times10^5\text{ J} = 200\text{ kJ}$

Practice Question

Q: A $0.20\text{ kg}$ ball moves at $15\text{ m/s}$ . Find its kinetic energy. What happens to the kinetic energy if its speed doubles?

Show Answer

$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.20)(15)^2 = 22.5\text{ J}$

Because $KE \propto v^2$ , doubling the speed quadruples the kinetic energy to $90\text{ J}$ .

Power

Power is how fast work is done, or energy is transferred, measured in watts (W). For a steady force moving an object at speed $v$ , power can also be written as $\;P = Fv$ .

Power

P = \frac{W}{t} = Fv

Worked Example

A motor does $60{,}000\text{ J}$ of work in $4.0\text{ s}$ . What is its power output?

Solution:

$P = \frac{W}{t} = \frac{60{,}000}{4.0} = 15{,}000\text{ W} = 15\text{ kW}$

Practice Question

Q: A car engine provides a steady driving force of $500\text{ N}$ while the car cruises at $30\text{ m/s}$ . What power is the engine delivering?

Show Answer

$P = Fv = (500)(30) = 15{,}000\text{ W} = 15\text{ kW}$

Pillar 3 of 3

Momentum, Energy and Simple Systems

Momentum

Momentum measures how hard it is to stop a moving object. It is a vector, the product of mass and velocity, measured in $\text{kg·m/s}$ .

Momentum

p = mv

Worked Example

Find the momentum of a $1500\text{ kg}$ car travelling at $20\text{ m/s}$ .

Solution:

$p = mv = (1500)(20) = 3.0\times10^4\text{ kg·m/s}$

Practice Question

Q: A $0.15\text{ kg}$ cricket ball travels at $40\text{ m/s}$ . Find its momentum. A $60\text{ kg}$ cyclist has the same momentum, how fast are they moving?

Show Answer

$p = mv = (0.15)(40) = 6\text{ kg·m/s}$

$v = \frac{p}{m} = \frac{6}{60} = 0.1\text{ m/s}$

A light, fast ball can carry the same momentum as a heavy, slow rider.

Impulse

J = Ft = \Delta p = m\Delta v

Worked Example

A constant force of $200\text{ N}$ acts on a ball for $0.50\text{ s}$ . Find the impulse delivered, and hence the change in the ball's momentum.

The impulse is the area under the force-time graph: 200 N × 0.5 s = 100 N·s.

Solution:

$J = Ft = (200)(0.50) = 100\text{ N·s}$

$\Delta p = J = 100\text{ kg·m/s}$

Practice Question

Q: A $0.058\text{ kg}$ tennis ball is served from rest to $30\text{ m/s}$ . Find the impulse. If the racket is in contact for $0.005\text{ s}$ , find the average force.

Show Answer

$J = m\Delta v = (0.058)(30) = 1.74\text{ N·s}$

$F = \frac{J}{t} = \frac{1.74}{0.005} = 348\text{ N}$

A longer contact time would lower this force, the same idea behind crumple zones.

Conservation of Momentum

In an isolated system (no external forces), the total momentum before an interaction equals the total momentum after it. This single principle solves almost every collision and explosion problem.

Conservation of Momentum

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Worked Example

A $2.0\text{ kg}$ trolley moving at $3.0\text{ m/s}$ collides with a stationary $1.0\text{ kg}$ trolley and they lock together. Find their common velocity afterward.

Momentum is conserved: 6 + 0 kg·m/s before equals 6 kg·m/s after, as the 2 kg trolley locks onto the stationary 1 kg trolley to move off at 2 m/s.

Solution (they stick, so share one final velocity):

$m_1u_1 + m_2u_2 = (m_1 + m_2)v$

$(2.0)(3.0) + 0 = (3.0)v$

$v = \frac{6.0}{3.0} = 2.0\text{ m/s}$

Practice Question

Q: A $60\text{ kg}$ astronaut floats at rest and throws a $2.0\text{ kg}$ tool at $5.0\text{ m/s}$ . Find the astronaut's recoil velocity.

Show Answer

Total momentum starts at zero and is conserved:

$0 = (2.0)(5.0) + (60)v$

$v = -\frac{10}{60} = -0.17\text{ m/s}$

The astronaut drifts at $0.17\text{ m/s}$ in the opposite direction to the tool.

Elastic and Inelastic Collisions

Momentum is conserved in every collision. Kinetic energy is not: whether it is conserved separates the two types.

Elastic

Both momentum and kinetic energy are conserved. Objects bounce apart cleanly (idealised, like billiard balls).

Inelastic

Momentum is conserved but kinetic energy is not; some is lost to heat, sound and deformation. If the objects stick together it is perfectly inelastic.

Worked Example

Compare kinetic energy before and after:

$KE_{\text{before}} = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\text{ J}$

$KE_{\text{after}} = \tfrac{1}{2}(3.0)(2.0)^2 = 6.0\text{ J}$

$\Delta KE = 6.0 - 9.0 = -3.0\text{ J}$

Momentum was conserved, but $3.0\text{ J}$ of kinetic energy became heat, sound and deformation, so the collision is inelastic.

Practice Question

Q: Two lumps of clay collide head-on and stick together. Is momentum conserved? Is kinetic energy conserved?

Show Answer

Momentum: yes, it is always conserved when no external forces act.

Kinetic energy: no. This is a perfectly inelastic collision, so some kinetic energy becomes heat and permanent deformation of the clay.

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