# Question 39b14

Nov 8, 2017

(a) Given street incline $= {20.0}^{\circ}$
Let Distance moved by Chemistro along the street$= l$
$\therefore$ the vertical height reached by Chemistro $h = l \sin {20}^{\circ}$

Total energy of Chemistro at any instant of time $= P E + K E$

$= m g h + \frac{1}{2} m {v}^{2}$
Where $m$ is mass of Chemistro and $g = 10 m {s}^{-} 2$ acceleration due to gravity

Initial total energy $= 0 + \frac{1}{2} \times 90.0 \times {\left(22.0\right)}^{2} = 21780 J$
Initial kinetic energy gets converted in $P E$ of reaching at height $h$ and rest is spent in doing work done against friction.

Final total energy when he comes to rest $= 90.0 \times 10 \times l \sin 20$ $= 900 l \sin {20}^{\circ} J$

Now work done by friction
$W = \text{Force of Friction"xx"Distance moved up} = \vec{{F}_{f}} \cdot \vec{d}$

From the figure we see that the normal reaction

$N = m g \cos {20}^{\circ}$

therefore force of the sliding friction is

F_f= μ_kN = μ_kmgcos20^@
:.W=-μ_kmgcos20^@xxl#
As force of friction and distance moved are opposite to each other we have $- v e$ sign.

Inserting given values and using law of conservation of energy we get

$21780 = 900 l \sin {20}^{\circ} + 0.333 \times 90.0 \times 10 l \cos {20}^{\circ}$
$\implies 21780 = 900 l \times 0.3420 + 0.333 \times 90.0 \times 10 l \times 0.9397$
$\implies 21780 = \left(307.8 + 281.6\right) l$
$\implies l = 37.0 m$, three significant figures
$\therefore h = 36.95 \times \sin {20}^{\circ}$
$\implies h = 12.6 m$

(b) Heat generated by sliding is same as work done against friction

$= 0.333 \times 90.0 \times 10 \times l \cos {20}^{\circ}$
$= 0.333 \times 90.0 \times 10 \times 36.95 \times 0.9397 = 10406 J$

(c) We know that power is the rate of doing work per unit time.

$\therefore$ Average power delivered by Power Man $= \frac{21780}{0.300}$ $= 72600 J {s}^{-} 1$