Kinetic Energy | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Kinetic Energy

Kinetic energy is the energy associated with motion.

Speed matters twice: if speed doubles while mass stays fixed, kinetic energy becomes four times as large.

Read the squared speed before calculating

E_k = \frac12 mv^2

Symbols

$E_k$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = speed (m s^{-1})

Change	Effect on kinetic energy
mass doubles; speed fixed	kinetic energy doubles
speed doubles; mass fixed	kinetic energy becomes four times as large
speed triples; mass fixed	kinetic energy becomes nine times as large

Worked example

A change in speed

A 800 kg car speeds up from 20 m s⁻¹ to 30 m s⁻¹. Find the change in kinetic energy.

Initial: $E_{k1}=\tfrac12(800)(20)^2=1.60\times10^5\text{ J}$ .
Final: $E_{k2}=\tfrac12(800)(30)^2=3.60\times10^5\text{ J}$ .
$\Delta E_k=E_{k2}-E_{k1}=2.00\times10^5\text{ J}$ .

Answer

2.00\times10^5\text{ J}

Common mistake

In general,

\Delta E_k\ne\tfrac12m(\Delta v)^2

. Calculate the two kinetic energies, then subtract.

Derive kinetic energy from work done

The work done by the resultant force equals the change in kinetic energy. This is different from the work done by one individual force when several forces act.

A resultant force accelerates a mass from rest, so $F=ma$ and $W=Fs=mas$ .
From $v^2=u^2+2as$ with $u=0$ , $as=\tfrac12v^2$ .
Substitute: $W=m(as)=\tfrac12mv^2$ . This work becomes the object's kinetic energy.

Your turn9702/22/M/J/25 Q1(b)(i)

A car starts from rest and accelerates with constant acceleration $a$ . After displacement $s$ , its speed is $v$ . Using work done, show that $E_k=\tfrac12mv^2$ . (9702/22/M/J/25 Q1(b)(i))

Show worked answer

Worked answer: $W=Fs=mas$ . Since $v^2=2as$ , $as=v^2/2$ . Therefore $E_k=W=m(v^2/2)=\tfrac12mv^2$ .