Chapter 6, Problem 63P

To determine

The power (watts and hp) tokeep the car travelling at a constant $80\text{km}/\text{h}$ .

Answer to Problem 63P

Solution:$38.29\text{hp}$ or $2.86\times {10}^4\text{W}$

Explanation of Solution

Given:

Mass of car $=1080\text{ kg}$

Final speed $=65\text{km}/\text{h}$

Initial speed $=95\text{km}/\text{h}$

Time $=7.0\text{s}$

FormulaUsed: From work-energy principle:

  $W_{net}=\Delta K=\left(\frac{1}{2}m{V}_2{}^2-\frac{1}{2}m{V}_1{}^2\right)$

Power $=\frac{W_{net}}{t}$

Calculation:

Convert speed from km/h to m/s:

  $V_1=\frac{95\times 1000\text{m}}{60\times 60\text{ s}}=26.39\text{m/s}$

  $V_2=\frac{65\times 1000\text{m}}{60\times 60\text{ s}}=18.06\text{m/s}$

  $W_{net}=\frac{1}{2}\times 1080\left[{\left(18.06\right)}^2-{\left(26.39\right)}^2\right]$

  $=540\left({\left(18.06\right)}^2-{\left(26.39\right)}^2\right)$

  $=-199944.99$

˜  ≈$-199945\text{ J}$

Negative work done means that work done by resistive forces opposes the motion.

Power of resistive force

  $\begin{array}{c}P=\frac{W_{net}}{t}\\ =\frac{199945\text{ J}}{7.0\text{ s}}\\ =28563.57\text{W}\end{array}$

Average power of resistive forces $=28563.57\text{ W}$

  $\left(746\text{ watt}=1\text{ hp}\right)$

  $=\frac{28563.57\text{ W}}{746\text{ W/hp}}$

  $=38.29\text{ hp}$

Resistive force is directly proportional to sped but $80\text{km}/\text{h}$ is average of $65\text{ km}/\text{h}$ and $95\text{km}/\text{h}$ So average resistive force between $65\text{km}/\text{h}$ to $95\text{km}/\text{h}$ , may be taken as resistive force at $80\text{km}/\text{h}$ .Power needed to keep the car travelling at a constant $\frac{80\text{km}}{\text{h}}=38.29\text{hp or }2.86\times {10}^4\text{ W}.$

Conclusion:

If car is moving with constant speed, then resistive force (friction, air drag etc) opposes the motion. So, to maintain steady speed one needs to supply energy.