JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 19)
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 :
20 ms$$-$$1
15 ms$$-$$1
10 ms$$-$$1
5 ms$$-$$1
Explanation
Here, tan$$\theta $$ = $${{100} \over {1000}} = {1 \over {10}}$$
$$ \therefore $$ sin$$\theta $$ = $${1 \over {10}}$$ (as $$\theta $$ is very small),
when car is moving uphill :
_9th_April_Morning_Slot_en_19_1.png)
P = f $$ \times $$ u
= (wsin$$\theta $$ + f) $$ \times $$ u
= $$\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$$
P = $${{3w} \over {20}} \times 10$$ = $${{3w} \over 2}$$
When car is moving down hill :
_9th_April_Morning_Slot_en_19_2.png)
$$ \therefore $$ $${P \over 2}$$ = (wsin$$\theta $$ $$-$$ f) $$ \times $$ v
$$ \Rightarrow $$ $${{3w} \over 4}$$ = $$\left( {{w \over {10}} - {w \over {20}}} \right)$$ $$ \times $$ v
$$ \Rightarrow $$ $${{w \over {20}} \times }$$ v = $${{3w} \over 4}$$
$$ \Rightarrow $$ v = 15 m/s
$$ \therefore $$ sin$$\theta $$ = $${1 \over {10}}$$ (as $$\theta $$ is very small),
when car is moving uphill :
P = f $$ \times $$ u
= (wsin$$\theta $$ + f) $$ \times $$ u
= $$\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$$
P = $${{3w} \over {20}} \times 10$$ = $${{3w} \over 2}$$
When car is moving down hill :
$$ \therefore $$ $${P \over 2}$$ = (wsin$$\theta $$ $$-$$ f) $$ \times $$ v
$$ \Rightarrow $$ $${{3w} \over 4}$$ = $$\left( {{w \over {10}} - {w \over {20}}} \right)$$ $$ \times $$ v
$$ \Rightarrow $$ $${{w \over {20}} \times }$$ v = $${{3w} \over 4}$$
$$ \Rightarrow $$ v = 15 m/s
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