(a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 96.0 km/h?

m

(b) If, in actuality, a 750 kg car with an initial speed of 96.0 km/h is observed to coast up a hill to a height 19.0 m above its starting point, how much thermal energy was generated by friction?

I CANT FIGURE OUT THIS ANSWER (PART B) . PLEASE HELP _________J

(c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? (Explicitly show on paper how you follow the steps in the Problem-Solving Strategy for energy found on pages 159 and 160. Your instructor may ask you to turn in this work.)

N (down the slope)

Guest Dec 2, 2014

#1**0 **

If you move your eyes straight to the bottom you will see that I am not comfortable with what I have done!

So you should probably wait for someone else.

For part A

You would need to know the angle incline of the slope.

Because you need to have the vertical component of the initial velocity

and that would be 96sin(angle)

I suppose that you would use 2.5 degrees

$$V_y=96sin(2.5^0) \approx 4.187461187 km/hour$$

$${\frac{{\mathtt{4.187\: \!461\: \!187}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{60}}\right)}} = {\mathtt{1.163\: \!183\: \!663\: \!055\: \!555\: \!6}}$$

approx vertical velocity is 1.1632 m/s Initially ** u=1.1632**

**a = -9.8**m/s^2

Find height **(s)** when velocity = 0 **v=0**

Number 4 has u, a, v, and s so use that one

$$\\v^2=u^2+2as\\\\

0=1.16318^2+2*-9.8*s$$

$${\mathtt{0}} = {{\mathtt{1.163\: \!18}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{s}}\right) \Rightarrow {\mathtt{s}} = {\frac{{\mathtt{114\,603\,045}}}{{\mathtt{1\,660\,192\,226}}}} \Rightarrow {\mathtt{s}} = {\mathtt{0.069\: \!029\: \!985\: \!326\: \!530\: \!5}}$$

This is the vertical height that it will coast in metres.

Draw the triangle and you can work out the distance it will coast up the hill

$${\frac{{\mathtt{0.069}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2.5}}^\circ\right)}}} = {\mathtt{1.581\: \!865\: \!408\: \!209\: \!866\: \!1}}$$

**1.58 metres - That does not look right **

**HELP - What did I do wrong ??**

Melody
Dec 2, 2014

#2**+5 **

»You would need to know the angle incline of the slope

Not when the question asks merely how high the hill

All the car's KE is converted to gravitational potential energy

½mv² = mgh

∴ h = ½v² ÷ g

DONE

Guest Dec 2, 2014

#3**0 **

part (b)

It doesn't gain as much gravitational PE as it would if losses were absent

so friction losses = mgh_{ideal - }mgh_{actual}

Guest Dec 2, 2014

#5**+5 **

Thanks you anon

ok so

96km/hour

$${\frac{{\mathtt{96}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{60}}\right)}} = {\frac{{\mathtt{80}}}{{\mathtt{3}}}} = {\mathtt{26.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$$\begin{array}{rll}

KEnergy&=&gravitational\; Potential \;energy\\

0.5mv^2&=&mgh\\

h&=&0.5v^2/g\\

h&=&0.5*(26.\dot6)^2/9.8 \;m\\

h&=&0.5*711.111/9.8\;m\\

h&\approx&36.28m

\end{array}$$

Is this correct? It does sound better.

If this one is correct then what was wrong with my logic the first time?

Melody
Dec 2, 2014

#6**+5 **

Melody, your initial logic (the use of v^{2} = u^{2} + 2as) would work if you imagine the car going straight up:

.

Alan
Dec 2, 2014