Conservation of mechanical energyEdexcel International A Level Further Maths: Revision notes

A force is conservative if the work it does depends only on the start and end points, not the path. Gravity and the normal reaction from a smooth surface (which does no work) are examples. When only such forces do work, the total mechanical energy (kinetic plus potential) stays constant: $\tfrac12mu^2+mgh_1=\tfrac12mv^2+mgh_2.$ Mass cancels, so the speed depends only on the change of height: for a particle moving freely, $v^2=u^2+2g(h_1-h_2)$. A ball thrown at any angle from a given height lands with the same speed.

On a smooth slope the normal reaction does no work, so energy is conserved. A particle released from rest at height $h$ above the foot reaches the foot with $v=\sqrt{2gh}$, whatever the angle or length of the slope. For a slope of length $4$ m at $30^\circ$, $h=4\sin30^\circ=2$ m and $v=\sqrt{2(9.8)(2)}=6.26$ m s$^{-1}$. For a vertical throw with speed $u$ from height $h_0$, the greatest height above the ground is $h_0+\frac{u^2}{2g}$. Use the energy equation whenever you need speed at a given height and the direction does not matter.

With a constant resistance (friction, air resistance) some mechanical energy is converted to heat. The correct statement is $\text{initial energy}-\text{work done against resistance}=\text{final energy}.$ The work done against a constant resistance $R$ over a distance $d$ is $Rd$, with $d$ the distance travelled along the path. Example: a child of mass $30$ kg drops $3$ m down a $6$ m slide and reaches $5$ m s$^{-1}$. Energy lost $=882-375=507$ J, so $6R=507$ and $R=84.5$ N.

On a rough plane inclined at $\alpha$, resolve perpendicular to the plane to find $R=mg\cos\alpha$, so friction is $F=\mu R$. Up the slope through $d$: $\tfrac12mu^2=\tfrac12mv^2+mgd\sin\alpha+Fd$. Down the slope through $d$: $\tfrac12mv^2=\tfrac12mu^2+mgd\sin\alpha-Fd$. On a rough horizontal surface, friction is $\mu mg$ and it removes the kinetic energy: $\mu mgd=\tfrac12mv^2$, so $d=\frac{v^2}{2\mu g}$. Example: arriving at the foot of a smooth $2$ m-high slope with $39.2$ J and then crossing a rough surface with $F=7.84$ N: $d=\frac{39.2}{7.84}=5$ m. In general $d=\frac h\mu$, because the mass cancels.

(1) Draw the start and end positions and mark their heights. (2) Write kinetic and potential energy at each. (3) Include the work done against any resistance on the right. (4) Solve for the unknown, then take a square root if you have $v^2$. Use energy for speeds and distances, and $F=ma$ for accelerations and forces.