FM1 2.1 - Work-energy principle and energy conservation

You already know

Use the resultant force in Newton's second law. Resolve forces parallel and perpendicular to a plane; while a particle slides, friction has magnitude $\mu R$ and opposes its motion. These A-level Mathematics skills supply the forces; this lesson connects their work to energy.

Work and kinetic energy

The work done by a force measures its energy transfer through displacement. Work is a scalar and may be positive, negative or zero. A particle's kinetic energy is the energy associated with its motion; it is non-negative and depends on speed, not direction. Both quantities are measured in joules, $\mathrm{J}$, with $1\,\mathrm{J}=1\,\mathrm{N\,m}$.

Constant-force work and the work-energy principle

For a constant force of magnitude $F$, displacement of magnitude $s$, and angle $\alpha$ between them,

For a particle of constant mass $m$ and speed $v$,

The resultant work is the sum of the work done by all forces. For initial and final speeds $u$ and $v$, this is $\tfrac12mv^2-\tfrac12mu^2$.

Not in the formula booklet: learn it.

Worked example 1: adding positive and negative work

Question. A particle of mass $2\,\mathrm{kg}$ moves along a horizontal track, initially at $3\,\mathrm{m\,s^{-1}}$. A constant driving force of $10\,\mathrm{N}$ acts horizontally forwards and a constant resistance of $2\,\mathrm{N}$ opposes the motion. Find the speed after a displacement of $5\,\mathrm{m}$. The particle remains in contact with the track.

Plan. Add the work of the driving and resisting forces and equate it to the kinetic energy increase.

Why this step? Resistance does negative work. Weight and the normal reaction are perpendicular to the horizontal displacement, so each does zero work.

Answer. The final speed is $7\,\mathrm{m\,s^{-1}}$.

Check. The resultant force is $8\,\mathrm{N}$, so $a=4\,\mathrm{m\,s^{-2}}$. Independently, $v^2=3^2+2(4)(5)=49$.

Worked example 2: a force at an angle

Question. A particle of mass $2\,\mathrm{kg}$ moves along a horizontal track, initially at $3\,\mathrm{m\,s^{-1}}$. A constant force of $10\,\mathrm{N}$ pulls forwards at $60^\circ$ above the horizontal; a constant resistance of $2\,\mathrm{N}$ opposes the motion. Find the speed after a horizontal displacement of $5\,\mathrm{m}$. The particle remains in contact with the track.

Plan. Use the pull's horizontal component to calculate its work; retain the same resisting work.

Why this step? The displacement is horizontal. The pull's vertical component changes the normal reaction, but it makes no direct contribution to the work. The resistance here is given as $2\,\mathrm{N}$, rather than calculated from a coefficient of friction.

Answer. The final speed is $2\sqrt6\,\mathrm{m\,s^{-1}}$, approximately $4.90\,\mathrm{m\,s^{-1}}$.

Check. The horizontal resultant is $10\cos60^\circ-2=3\,\mathrm{N}$. Thus $v^2=9+2(3/2)(5)=24$, agreeing with the energy calculation.

Watch out

Wrong: $10(5)-2=\Delta K$. The first term is work in joules, but the second is force in newtons. Right: $10(5)-2(5)=\Delta K$ for the horizontal pull. Every work term needs the appropriate displacement.

Gravitational potential energy

The gravitational potential energy of a particle near the Earth's surface is $U=mgh$, where $h$ is its vertical height above a chosen reference level. Only differences in potential energy affect a motion calculation. A negative $h$, and hence a negative $U$, is allowed below the chosen level.

Gravitational energy and conservation

For constant $g$ and vertical height change $\Delta h$,

If gravity is the only force doing work,

Thus $K+U$, the mechanical energy in this model, is constant. A stationary smooth track's normal reaction does no work because it is perpendicular to motion. Use $g=9.8\,\mathrm{m\,s^{-2}}$.

Not in the formula booklet: learn it.

Worked example 3: ascending a smooth plane

Question. A particle of mass $2\,\mathrm{kg}$ moves up a smooth fixed plane with initial speed $7\,\mathrm{m\,s^{-1}}$. The plane is inclined at $\theta$, where $\sin\theta=0.4$. Find its speed after travelling $5\,\mathrm{m}$ up the plane. Ignore air resistance.

Plan. Subtract the gain in gravitational potential energy from the initial kinetic energy.

Why this step? The plane's reaction does no work. Kinetic energy supplies the rise in potential energy, so the speed decreases.

Answer. The speed is $3.13\,\mathrm{m\,s^{-1}}$.

Check. The energy remaining, $9.8\,\mathrm{J}$, is positive, so the particle can reach this point. The gain $39.2\,\mathrm{J}$ and remaining $9.8\,\mathrm{J}$ total the initial $49\,\mathrm{J}$.

Worked example 4: descending a smooth plane

Question. A particle of mass $2\,\mathrm{kg}$ moves down a smooth fixed plane with initial speed $7\,\mathrm{m\,s^{-1}}$. The plane is inclined at $\theta$, where $\sin\theta=0.4$. Find its speed after travelling $5\,\mathrm{m}$ down the plane. Ignore air resistance.

Plan. Add the gravitational potential energy lost to the initial kinetic energy.

Why this step? The potential energy change is negative, so subtracting it increases the kinetic energy. A loss of potential energy is not a loss of total mechanical energy.

Answer. The speed is $9.39\,\mathrm{m\,s^{-1}}$.

Check. With the starting height as zero, final potential energy is $-39.2\,\mathrm{J}$. Then final $K+U=88.2-39.2=49\,\mathrm{J}$, the initial total.

Watch out

Do not include both $-mg\Delta h$ as gravity's work and $+mg\Delta h$ as potential energy gained in the same balance. Either use the work of every force to calculate $\Delta K$, or move gravity into $\Delta U$. Counting it twice gives a false energy loss or gain.