# A car accelerates down a hill that is 45.0^@ to the horizontal. If the car has a 350kW engine and takes 2.4s to reach the bottom, what is the force of friction resisting its motion?

## A car of $750 k g$ accelerates down a hill. At the top of the hill, its speed is $75 \frac{k m}{h}$, while at the bottom, its speed is $185 \frac{k m}{h}$. The hill is $40.0 m$ long and makes an angle of ${45.0}^{\circ}$ with the horizontal. If the car has a $350 k W$ engine and takes $2.4 s$ to reach the bottom, what is the force of friction resisting its motion?

Jun 24, 2016

5508.575N

#### Explanation:

Given

• $m \to \text{Mass of the car} = 750 k g$
• ${v}_{i} \to \frac{\text{Initial velocity of the car at top of the hill"= 75"km}}{h}$
$= \frac{75 \times 1000}{3600} \frac{m}{s} = \frac{375}{18} \frac{m}{s}$
• ${v}_{f} \to \frac{\text{Final velocity of the car at hill bottom "= 185"km}}{h}$
$= \frac{185 \times 1000}{3600} \frac{m}{s} = \frac{925}{18} \frac{m}{s}$
• $d \to \text{ Length of the hill path} = 40 m$
• $a \to \text{Angle of inclination of path } = {45}^{\circ}$
• $P \to \text{The power of car engine} = 350 k W = 350 \times {10}^{3} W$
• $t \to \text{The time descent from top to bottom} = 2.4 s$
• F->"Force of friction (in N)"=?#

We are to apply the law of conservation of energy to solve the problem in the easiest way.

Here the net gain in KE of car will be equal to the sum of the enegy supplied by the engine and work done by the gravitational pull minus the loss of energy due to friction( i.e. work done against force of friction)

Let us calculate each separately

• $\text{Total gain in KE} , \Delta {E}_{k} = \frac{1}{2} m \left({v}_{f}^{2} - {v}_{i}^{2}\right)$

$= \frac{1}{2} \times 750 \times \left({\left(\frac{925}{18}\right)}^{2} - {\left(\frac{375}{18}\right)}^{2}\right) J$

$= 375 \times \frac{1300 \times 550}{18 \times 18} = 827546.3 J$

• ${E}_{\text{engine"->"Energy supplied by engine}} = P \times t$

${E}_{\text{engine}} = 350000 \times 2.4 J = 840000 J$

• ${W}_{g} \to \text{work done by gravitational pull} = m g \sin a \times d$

$= 750 \times 9.8 \times \sin 45 \times 40 J = 207889.3 J$

• ${W}_{\text{friction}} = F \times d = F \times 40 J$

So by the law of conservation of energy

$\Delta {E}_{k} = {E}_{\text{engine"+W_g-W_"friction}}$

$\implies {W}_{\text{friction"=E_"engine}} + {W}_{g} - \Delta {E}_{k}$

$\implies F \times 40 = 840000 + 207889.3 - 827546.3 = 220343$

$\therefore F = \frac{220343}{40} = 5508.575 N$