A cyclist and her bicycle have a combined mass of 60 kg. (a) Calculate the kinetic energy store of the cyclist and bicycle when travelling at a speed of 8.0 m/s. [2] (b) The cyclist speeds up to 16 m/s, which is double her original speed. Calculate the new kinetic energy store o…

Answer: ((a)) 1920 J ((b)) 7680 J ((c)) Doubling the speed multiplies the kinetic energy by 4 (not 2), because speed is squared in Ek = 1/2mv^2.. Part (a): Kinetic energy at 8.0 m/s Use Ek = 1/2mv^2 with m = 60 kg and v = 8.0 m/s: Ek = 1/2 × 60 × 8.0^2 Ek = 1/2 × 60 × 64 Ek = 30 × 64 = 1920 J Part (b): Kinetic energy at 16 m/s The speed is now v = 16 m/s: Ek = 1/2 × 60 × 16^2 Ek = 1…

IGCSE Physics 0625: Kinetic Energy and Doubling the Speed (Ek = ½mv²)

Part (a): Kinetic energy at 8.0 m/s

Use $E_k = \tfrac{1}{2}mv^2$ with $m = 60\text{ kg}$ and $v = 8.0\text{ m/s}$ :

$E_k = \tfrac{1}{2} \times 60 \times 8.0^2$

$E_k = \tfrac{1}{2} \times 60 \times 64$

$E_k = 30 \times 64 = 1920\text{ J}$

Part (b): Kinetic energy at 16 m/s

The speed is now $v = 16\text{ m/s}$ :

$E_k = \tfrac{1}{2} \times 60 \times 16^2$

$E_k = \tfrac{1}{2} \times 60 \times 256$

$E_k = 30 \times 256 = 7680\text{ J}$

Part (c): Effect of doubling the speed

Compare the two kinetic energies:

$\frac{7680}{1920} = 4$

Doubling the speed makes the kinetic energy 4 times as large, not twice as large. This is because $v$ is squared in the equation $E_k = \tfrac{1}{2}mv^2$ : if $v$ becomes $2v$ , then $v^2$ becomes $(2v)^2 = 4v^2$ , so $E_k$ becomes $4$ times its original value.

Final answers

(a) Kinetic energy at $8.0\text{ m/s}$ $=$ $1920\text{ J}$
(b) Kinetic energy at $16\text{ m/s}$ $=$ $7680\text{ J}$
(c) Doubling the speed quadruples the kinetic energy, because kinetic energy is proportional to $v^2$