A long stick rests on the surface. A person standing $ 10\;m $ away from the stick. With what minimum speed an object of mass $ 0.5\;kg $ should he have thrown so that it hits the stick. Assume the coefficient of kinetic friction is $ 0.7\; $

Hint: A person is standing $ 10\;m $ away from a stick that is lying on a surface. He wants to throw an object of mass $ 0.5\;kg $ at the stick. We have to find the minimum speed at which he should throw the object so that it will hit the stick. The value of the coefficient of kinetic friction is given as $ 0.7\; $ .

Complete Step by step solution:
The distance between the man and the stick is given as, $ d = 10m $
The mass of the object that is thrown by the man is given as $ m = 0.5kg $
The work done in moving the stick in a horizontal plane will be the product of the force applied and displacement of the stick. This can be written as,
 $ w = f \times d $
Here the force can be written as,
 $ f = \mu R $
Where $ \mu $ is the coefficient of kinetic friction and $ R $ stands for the normal reaction i.e. the component of force normal to the surface. This can be written as
  $ R = mg\cos \theta $
The stick is lying on the surface, therefore the angle $ \theta = 0 $
Therefore, $ R = mg $ because $ \cos 0 = 1 $
Then the force can be written as,
 $ f = \mu mg $
The work done can be written as,
 $ w = \mu mgd $
The work done will be the kinetic energy of the object when it is thrown.
i.e.
 $ \dfrac{1}{2}m{v^2} = \mu mgd $
From this, the velocity can be written as,
 $ {v^2} = \dfrac{{2\mu mgd}}{m} = 2\mu gd $
We know that
 $ d = 10m $
 $ \mu = 0.7 $
 $ g = 9.8m/{s^{}} $
Substituting the values, we get
 $ {v^2} = 2 \times 0.7 \times 9.8 \times 10 = 137.2 $
Taking the root, we get
 $ v = \sqrt {137.2} = 11.71m/s $
The answer is: $ 11.71m/s $ .

Note:
The friction experienced by a body when it moves is called kinetic friction. The kinetic friction has a constant value depending on the nature of the surfaces that are in contact. The kinetic friction is proportional to the normal reaction. The coefficient of kinetic friction is smaller than that of the coefficient of static friction.