- #1

AntSC

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## Homework Statement

a) A car of mass 1050kg moves along a straight horizontal road with its engine working at a constant rate of 25kW. Its speed at a point A on the road is 12ms

^{-1}. Assuming that there is no resistance to motion, calculate the time taken for the car to travel from A until it reaches a speed of 20ms

^{-1}.

b) Assume now that there is a constant resistance to motion and that the car's engine continues to work at 25kW. It takes 10.7s for the car's speed to increase from 12ms

^{-1}to 20ms

^{-1}. During this time the car travels 179m. Calculate the work done against resistance and hence find the magnitude of the resistance.

c) Later the car moves up a straight hill, inclined at 2° to the horizontal. The engine works at 25kW as before, and there is a constant resistance of the same magnitude as before. The car travels a distance of 393m while its speed increases from 12ms

^{-1}to 20ms

^{-1}. Calculate the time taken by the car to travel this distance.

## Homework Equations

I'm lost at part b). I don't know what i am missing with my understanding.

## The Attempt at a Solution

a)

Given that gravitational force is conservative, the work-energy principle can be used.

[tex]P=\frac{E}{t}=\frac{m\left ( v^{2}-u^{2} \right )}{2t}[/tex]

[tex]t=\frac{m\left ( v^{2}-u^{2} \right )}{2P}[/tex]

[tex]t=\frac{1050\left ( 20^{2}-12^{2} \right )}{2\times 25000}[/tex]

[tex]t=5.376s[/tex]

This answer is confirmed correct

b)

[itex]P=Fv[/itex] so [itex]F=\frac{P}{v}[/itex]. [itex]F=ma[/itex] so with a resistive force

[tex]\frac{P}{v}-R=ma[/tex]

Given that work done [itex]W=Fd[/itex] and the net force in the system is now [itex]\frac{P}{v}-R[/itex] then

[tex]W=\left ( \frac{P}{v}-R \right )d[/tex]

So we need the magnitude of [itex]R[/itex] before we can find work done. To find [itex]R[/itex] we need to know the acceleration.

[tex]a=\frac{v-u}{t}[/tex]

[tex]a=\frac{20-12}{10.7}[/tex]

[tex]a=0.748ms^{-2}[/tex]

Now can find [itex]R[/itex]

[tex]R=\frac{P}{v}-ma[/tex]

[tex]R=\frac{25000}{12}-1050\left ( 0.748 \right )[/tex]

[tex]R=1297.9N[/tex]

This is incorrect. Don't know what I'm missing here. Any help would be great. Cheers