Work, Energy and Power: Question 10

Worked solution

Part (a): Initial kinetic energy

Using $E_k=\tfrac12mv^2$ with the speed before braking: $E_k = \tfrac12\times1200\times20^2$

Squaring the speed first: $20^2=400$. Multiplying by the mass: $1200\times400=480000$. Halving: $\tfrac12\times480000=240000\text{ J}$.

Check by recomputing in a different order: $\tfrac12\times1200=600$, and $600\times400=240000\text{ J}$. Both routes agree.

So the initial kinetic energy is $\boxed{240000}\text{ J}=2.4\times10^{5}\text{ J}$.

Part (b): Braking force

By the work–energy principle, the work done by the (constant) braking force equals the loss in kinetic energy of the car. Since the car comes to rest, its final kinetic energy is zero, so the entire initial kinetic energy found in part (a) is removed: $Fs = E_k$

Rearranging for the braking force $F$, using the braking distance $s=50\text{ m}$: $F = \frac{E_k}{s} = \frac{240000}{50} = 4800\text{ N}$

Check: $4800\times50=240000\text{ J}$, which matches the initial kinetic energy from part (a). Confirmed.

So the braking force is $\boxed{4800}\text{ N}$.

Part (c): An additional assumption

The calculation in part (b) also assumes that no other resistive forces (such as air resistance) act on the car during braking, so that the braking force alone accounts for the entire loss of kinetic energy. (Alternatively: that the road is horizontal, so no component of the car’s weight acts along the direction of motion.)

Final answers

• (a) Initial kinetic energy $= \boxed{240000}\text{ J} = 2.4\times10^{5}\text{ J}$
• (b) Braking force $= \boxed{4800}\text{ N}$
• (c) Assumption: no other resistive forces (e.g. air resistance) act, and/or the road is level