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(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

User Bumsoverboard
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1 Answer

23 votes
23 votes

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}*\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=(1)/(2)mv^2

Substituting all known values,


\begin{gathered} KE=(1)/(2)*(1000\text{ kg})*(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)=(1)/(2)mv^(\prime2) \\ =(1)/(2)*(1000\text{ kg})*(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.

User Jelmer Keij
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