Question 160: Work, Energy and Power & Circular Motion

Question

A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km
and applies a constant frictional force $${W \over 20}$$ on the car. While moving uphill on the road at a speed of 10 ms−1, the car needs power P. If it needs power $${p \over 2}$$ while moving downhill at speed v then value of $$\upsilon $$ is :

Accepted Answer

Here, tan$$	heta $$ = $${{100} \over {1000}} = {1 \over {10}}$$
$$ 	herefore $$   sin$$	heta $$ = $${1 \over {10}}$$ (as   $$	heta $$  is very small),
when car is moving uphill :
P = f $$ 	imes $$ u
=  (wsin$$	heta $$ + f) $$ 	imes $$ u
=  $$\left( {{w \over {10}} + {w \over {20}}} ight) 	imes 10$$
P  =  $${{3w} \over {20}} 	imes 10$$ = $${{3w} \over 2}$$
When car is moving down hill :
$$ 	herefore $$   $${P \over 2}$$ = (wsin$$	heta $$ $$-$$ f) $$ 	imes $$ v
$$ \Rightarrow $$   $${{3w} \over 4}$$ = $$\left( {{w \over {10}} - {w \over {20}}} ight)$$ $$ 	imes $$ v
$$ \Rightarrow $$   $${{w \over {20}} 	imes }$$ v = $${{3w} \over 4}$$
$$ \Rightarrow $$   v = 15 m/s

Answer

20 ms$$-$$1

Answer

15 ms$$-$$1

Answer

10 ms$$-$$1

Answer

5 ms$$-$$1

Work, Energy and Power & Circular Motion

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