Question

(a) . what is the kinetic energy in joules of a 1000 kg automobile traveling and 90 km/h(b) . how much work would have to be done to bring a 1000 kg automobile traveling at 90 km/h to a stop

Answer 1

Part (a)

Given data:

Mass of the automobile;

$m=1000\text{ kg}$

Velocity of the automobile;

$v=90\text{ km/h}$

Converting the velocity from km/h to m/s:

$\begin{gathered} v=90\text{ km/h}\times\frac{1000\text{ m}}{3600\text{ s}} \\ =25\text{ m/s} \end{gathered}$

The kinetic energy of the automobile is given as,

$K\mathrm{}E=\frac{1}{2}mv^2$

Substituting all known values,

$\begin{gathered} KE=\frac{1}{2}\times(1000\text{ kg})\times(25\text{ m/s})^2 \\ =312500\text{ J } \end{gathered}$

Therefore, kinetic energy of the automobile is 312500 J.

Part (b)

When the automobile is stoped, its velocity becomes 0 i.e. v'=0. Therefore, kinetic energy of the automobile when it comes to stop is given as,

$\begin{gathered} K\mathrm{}E^{\prime}=\frac{1}{2}mv^{\prime2} \\ =\frac{1}{2}\times(1000\text{ kg})\times(0\text{ m/s})^2 \\ =0\text{ J} \end{gathered}$

According to work-energy theorem, the work is equal to change in kinetic energy. Therefore,

$W=K\mathrm{}E^{\prime}-K\mathrm{}E\text{. }$

Substituting all known values,

$\begin{gathered} W=0\text{ J}-312500\text{ J} \\ =-312500\text{ J} \end{gathered}$

Here, negative sign indicates that the work is done on the system.

Therefore, the work done to bring the automobile to stop is 312500 J.