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# (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?

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(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?
(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)

physics
Guest Dec 2, 2014

#8
+26399
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Alan  Dec 2, 2014
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#1
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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 = -9.8m/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

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

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 slopeNot when the question asks merely how high the hillAll the car's KE is converted to gravitational potential energy
½mv² = mgh
∴ h = ½v² ÷ gDONE










































Guest Dec 2, 2014
#3
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part (b)

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

so friction losses = mghidealmghactual

Guest Dec 2, 2014
#4
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I'm not sure what I did to get Courier text and all those blank lines.

# ☹

Guest Dec 2, 2014
#5
+91436
+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
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Melody, your initial logic (the use of v2 = u2 + 2as) would work if you imagine the car going straight up:

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Alan  Dec 2, 2014
#7
+91436
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Thanks Alan,

I thought I was just taking the upward component.  I cannot do that I am supposing. :)

Melody  Dec 2, 2014
#8
+26399
+5

.

Alan  Dec 2, 2014
#9
+91436
0

Thanks Alan

Melody  Dec 2, 2014

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