A stone of 1 kg is thrown with a velocity of $20\ ms^{-1}$ across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?

AcademicPhysicsNCERTClass 9

Given:

A stone of $1\ kg$ is thrown with a velocity of $20\ ms^{-1}$ across the frozen surface of a lake and comes to rest after traveling a distance of $50\ m$. 


To do:

To find the force of friction between the stone and the ice.

Solution:

When a stone is thrown with a velocity of $20\ ms^{-1}$ across the frozen surface of a lake and comes to rest after traveling a distance of $50\ m$.  Friction offered by the frozen surface of the lake causes the stone to come to rest.



Calculation of friction:

Here, the mass of the stone$=1\ kg.$

When the stone is thrown, the initial velocity of the stone, $u=20\ ms^{-1}$

When the stone comes to rest, the final velocity of the stone, $v=0$

Distance covered by the stone before coming to the rest, $s=50\ m$

On using the second equation of motion, $v^2=u^2+2as$

$0^2=20^2+2\times a\times 50$

Or $0=400+100a$

Or $100a=-400$

Or $a=-\frac{400}{100}$

Or $a=-4\ ms^{-2}$                   [$-ve$ sign indicates the retardation]

Therefore, friction between the stone and the ice causes the stone to come to rest.

Let us calculate the friction offered by the frozen surface of the lake to the stone by using Newton's second law of motion.

$\boxed{F=ma}$

Friction offered by the lake's surface $=1\ kg. \times (-4\ ms^{-2})$

$=-4\ N$

Conclusion:

Therefore, friction offered by the frozen surface of the lake is $-4\ N$.

Here, $-ve$ sign indicates the opposing force$(friction)$.

raja
Updated on 10-Oct-2022 13:22:30

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