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

Last updated date: 17th Apr 2024
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Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

Before we proceed, we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force acting in leftward while the stone is moving rightward direction. One thing we have to fix in our mind that frictional force will always act opposite to the direction of motion of the body. Now on resolving the forces, $\mathrm{f}-\mathrm{ma} = 0$ or $\mathrm{f}=\mathrm{ma}$ also form the equation of motion $\mathrm{v}^{2}=\mathrm{u}^{2}+2as$ as, here final velocity is zero, so $\mathrm{v}=0$
$\mathrm{u}^{2}=-2 \mathrm{as}$
$\mathrm{u}^{2}=-2 \mathrm{as}$
$\mathrm{a}=\dfrac{-\mathrm{u}^{2}}{2 \mathrm{s}}=-\dfrac{20^{2}}{2 \times 50}=-4 \dfrac{\mathrm{m}}{\mathrm{sec}^{2}}$
On substituting the given data in equation (1) $\mathrm{f}=1 \times(-4)=-4 \mathrm{N}$