Question

A 2600 kg truck travelling at 72 km/h slams on the brakes and skids to a stop. The frictional force from the road is 8200 N. Use the relationship between kinetic energy and mechanical work to determine the distance it takes for the truck to stop.

Answer 1

Approximately
`63\; \rm m`.

Step-by-step explanation:

Convert the initial speed of this truck to standard units:

`\begin{aligned} v &= 72\; \rm km \cdot h^(-1) * (1\; \rm h)/(3600\; \rm s) * (1000\; \rm m)/(1\; \rm km) \\ &= 20\; \rm m \cdot s^(-1)\end{aligned}`.

Calculate the initial kinetic energy of this truck:

`\begin{aligned}\text{KE} &= (1)/(2) \, m \cdot v^(2) \\ &= (1)/(2) * 2600\; \rm kg * (20\; \rm m \cdot s^(-1))^(2) \\ &= 5.2 * 10^(5) \; \rm J\end{aligned}`.

The kinetic energy of the truck would be
`0` when the truck is not moving. Hence, the friction on the truck would need to do
`(-5.2 * 10^(5)\; \rm J)` of work on the truck to bring the truck to a stop.

Calculate the displacement required to achieve
`(-5.2 * 10^(5)\; \rm J)` of work at
`8200\; \rm N`:

`\begin{aligned}x &= (W)/(F) \\ &= (-5.2 * 10^(5)\; \rm J)/(8200\; \rm N) \approx -63\; \rm m\end{aligned}`.

This result is negative because the direction of friction is opposite to the direction of the motion of the truck.

Hence, the truck would come to a stop after skidding for approximately
`63\; \rm m`.