JEE MAIN - Physics (2005 - No. 60)
Consider a car moving on a straight road with a speed of $$100$$ $$m/s$$. The distance at which car can be stopped is $$\left[ {{\mu _k} = 0.5} \right]$$
$$1000$$ $$m$$
$$800$$ $$m$$
$$400$$ $$m$$
$$100$$ $$m$$
Explanation
Acceleration due to friction = $$\left( { - {\mu _k}g} \right)$$
We know, $${v^2} = {u^2} + 2as$$
$$ \Rightarrow $$ $${0^2} = {u^2} + 2\left( { - {\mu _k}g} \right)s$$
$$ \Rightarrow $$ $$2 { {\mu _k}g}s$$ = $${u^2}$$
$$ \Rightarrow s = {{{{100}^2}} \over {2 \times 0.5 \times 10}}$$
$$ \Rightarrow s = 1000\,m$$
We know, $${v^2} = {u^2} + 2as$$
$$ \Rightarrow $$ $${0^2} = {u^2} + 2\left( { - {\mu _k}g} \right)s$$
$$ \Rightarrow $$ $$2 { {\mu _k}g}s$$ = $${u^2}$$
$$ \Rightarrow s = {{{{100}^2}} \over {2 \times 0.5 \times 10}}$$
$$ \Rightarrow s = 1000\,m$$
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