Dynamics, Work and Energy — IMAT Physics Formulas and Traps

Mass, Weight and Newton's Three Laws

Mass and weight are different quantities

Weight

W = mg

$W$ weight, in N
$m$ mass, in kg
$g$ gravitational field strength, in N/kg

Newton's second law: resultant force equals mass times acceleration

Newton's second law

F_{\text{net}} = ma

$F_{\text{net}}$ resultant of all forces on the body, in N
$m$ mass, in kg
$a$ acceleration, in m/s²

Newton's third law: action and reaction pairs

Force	Its third-law partner	Type
Earth pulls a book down (its weight)	The book pulls the Earth up	gravitational
Table pushes up on the book	Book pushes down on the table	contact (normal)
Your foot pushes the ground backwards	The ground pushes your foot forwards	friction
A rocket pushes exhaust gas backwards	The gas pushes the rocket forwards	contact
A swimmer pushes water backwards	The water pushes the swimmer forwards	contact

Common traps

The kilogram is not a unit of weight

In everyday speech people weigh 60 kg, but in physics weight is a force in newtons. A 60 kg person weighs about 590 N on Earth. An option giving a weight in kg, or a mass in N, is wrong.

Motion does not need a resultant force

A body moving at constant velocity has zero resultant force. A resultant force is needed to change the velocity, not to keep it. Options claiming that the forward force must exceed the drag for a car to keep moving steadily are wrong.

Subtract the weight for anything pushed upwards

A rocket's thrust

T

is not its resultant force: its weight

mg

acts the other way. The starting acceleration is

a = T/m - g

, not

T/m

. Writing

T/m - mg

mixes an acceleration with a force and cannot be right.

At the top of a vertical throw, the weight still acts

The ball is momentarily at rest, but gravity has not switched off. The resultant force is the weight, and the acceleration is

g

downwards, the same as at every other point of the flight.

Weight and the normal force are not a third-law pair

Both act on the same body, and they are different kinds of force. They are equal only when the body is not accelerating vertically. The real partner of a body's weight is its gravitational pull on the Earth.

Friction, Inclined Planes, Circular Motion and Gravitation

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Static and kinetic friction

Friction

f_s \le \mu_s N \qquad f_k = \mu_k N

$\mu_s, \mu_k$ coefficients of static and kinetic friction (no unit)
$N$ normal reaction force, in N

Forces on an inclined plane

Sliding down a slope

N = mg\cos\theta \qquad ma = mg\sin\theta - \mu_k mg\cos\theta

$\theta$ angle of the slope to the horizontal
$N$ normal reaction from the slope

Centripetal force for motion in a circle

Centripetal force

F = \frac{mv^2}{r} = m\omega^2 r

$m$ mass, in kg
$v$ speed, in m/s
$r$ radius of the circle, in m

Newton's law of universal gravitation

Gravitational force and field strength

F = \frac{G m_1 m_2}{r^2} \qquad g = \frac{GM}{r^2}

$G$ gravitational constant, about 6.67 × 10⁻¹¹ N m² kg⁻²
$m_1, m_2, M$ masses, in kg
$r$ distance between centres, in m

Common traps

Static friction is only as big as it needs to be

\mu_s N

is the largest value static friction can reach, not its value. A body at rest under a small push feels a friction equal to that push. Use

\mu N

only when the body is sliding or just about to slide.

On a slope the normal reaction is less than the weight

The surface only has to balance the part of the weight pressing into it,

mg\cos\theta

. Using

N = mg

on a slope makes the friction too large.

Uniform circular motion has a non-zero resultant force

The speed is constant, but the resultant force is

mv^2/r

towards the centre. An option giving zero resultant force, or a force along the direction of motion, is wrong.

Measure r from the centre, not from the surface

A body at a height equal to two Earth radii above the surface is three radii from the centre, so

g

there is

g/9

, not

g/4

. Add the radius to the height before squaring.

Momentum, Impulse and Collisions

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Momentum and impulse

p = mv \qquad F\,\Delta t = \Delta p = mv - mu

$p$ momentum, in kg m/s
$F$ average force, in N
$\Delta t$ time the force acts, in s

Conservation of momentum in collisions and explosions

Conservation of momentum

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

$u_1, u_2$ velocities before the collision, with signs
$v_1, v_2$ velocities after the collision, with signs

Common traps

A rebound changes the momentum by more than a stop does

Momentum is a vector. A ball arriving at 20 m/s and leaving at 30 m/s the other way changes its velocity by 50 m/s, not 10 m/s. Bouncing back always needs a bigger impulse than simply stopping.

Momentum is always conserved; kinetic energy only in elastic collisions

In any collision with no external force the total momentum stays the same. Kinetic energy is conserved only when the collision is elastic. An option saying both are always conserved, or that momentum is lost when bodies stick together, is wrong.

Velocities after a collision carry signs

A light body bouncing off a heavier one at rest moves backwards. Writing every speed as positive in the momentum equation gives an answer that breaks conservation of momentum.

Work, Kinetic Energy, Potential Energy and Springs

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Work done by a force

Work

W = Fs\cos\theta

$F$ size of the force, in N
$s$ displacement, in m
$\theta$ angle between force and displacement

Kinetic and gravitational potential energy, and the work-energy theorem

Kinetic and potential energy

E_k = \tfrac{1}{2}mv^2 \qquad \Delta E_p = mg\,\Delta h \qquad W_{\text{net}} = \Delta E_k

$m$ mass, in kg
$v$ speed, in m/s
$\Delta h$ change of height, in m

Hooke's law and elastic potential energy

Hooke's law and stored energy

F = kx \qquad E_{\text{el}} = \tfrac{1}{2}kx^2

$k$ spring constant, in N/m
$x$ extension or compression, in m

Common traps

No movement along the force means no work

A force that is perpendicular to the motion, or that moves nothing, does zero work, however large it is. Effort and tiredness are not work in the physics sense.

Kinetic energy goes with the square of the speed

Doubling the speed multiplies the kinetic energy by four, and a speed left in km/h makes the answer

3.6^2 \approx 13

times too large. Convert to m/s first, then square.

Stored energy is half force times extension, not force times extension

The force grows from zero to

kx

during the stretch, so the work done is the area of a triangle:

\tfrac{1}{2}kx^2

. Using

Fx

with the final force doubles the answer.

Conservation of Mechanical Energy, Power and Efficiency

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Conservation of mechanical energy, and what friction does

Mechanical energy without friction

\tfrac{1}{2}mv^2 + mgh = \text{constant} \qquad v = \sqrt{2gh}\ \text{(from rest)}

$h$ height, in m
$v$ speed, in m/s

Power and efficiency

P = \frac{W}{t} = Fv \qquad \eta = \frac{\text{useful output}}{\text{total input}} \times 100\%

$P$ power, in W
$W$ work done or energy transferred, in J
$\eta$ efficiency

Common traps

Friction lowers mechanical energy but not total energy

On a rough slope the mechanical energy of a sliding body decreases; it does not stay constant, double or vanish. The missing amount has become heat, so the total energy is still conserved.

The kilowatt-hour is energy, not power

A kilowatt is a rate; a kilowatt-hour is an amount of energy,

3.6 \times 10^6\ \text{J}

. Likewise a power needs a time or a speed: a force and a distance alone give work, not power.

Moments, Couples, Equilibrium and Centre of Mass

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The moment of a force, and couples

Moment and couple

M = F d_{\perp} \qquad \tau_{\text{couple}} = F d

$d_{\perp}$ perpendicular distance from the pivot to the line of the force, in m
$d$ perpendicular distance between the two forces of a couple, in m

Equilibrium and the principle of moments

Principle of moments

\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}} \qquad \sum F_{\text{up}} = \sum F_{\text{down}}

$M$ moment about the chosen point, in N m

Centre of mass and stability

Centre of mass of two masses

x_{\text{cm}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

$x_1, x_2$ positions of the two masses along the line
$m_1, m_2$ the two masses

Common traps

A couple has zero resultant force but a non-zero torque

Two equal and opposite forces on different lines cancel as forces, so the body does not accelerate along, but they still turn it. The torque is one force times the distance between the lines, not zero and not twice that.

Include the beam's own weight, at its centre

Unless the beam is called light or of negligible mass, its weight acts at its midpoint and has a moment like any other load. Leaving it out, or placing it at the pivot, gives the wrong balance.

The centre of mass is nearer the heavier mass

The heavier mass pulls the balance point towards itself. Using the mass ratio the wrong way round puts the centre of mass near the light end, and that reversed answer is usually one of the options.

Dynamics, Work and Energy formulas

Mass, Weight and Newton's Three Laws

Mass and weight are different quantities

Newton's second law: resultant force equals mass times acceleration

Newton's third law: action and reaction pairs

Common traps

Friction, Inclined Planes, Circular Motion and Gravitation

Static and kinetic friction

Forces on an inclined plane

Centripetal force for motion in a circle

Newton's law of universal gravitation

Common traps

Momentum, Impulse and Collisions

Conservation of momentum in collisions and explosions

Common traps

Work, Kinetic Energy, Potential Energy and Springs

Work done by a force

Kinetic and gravitational potential energy, and the work-energy theorem

Hooke's law and elastic potential energy

Common traps

Conservation of Mechanical Energy, Power and Efficiency

Conservation of mechanical energy, and what friction does

Common traps

Moments, Couples, Equilibrium and Centre of Mass

The moment of a force, and couples

Equilibrium and the principle of moments

Centre of mass and stability

Common traps

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