a stone of 1kg is thrown with a velocity of 20m/s. across a frozen surface of a lake and comes to rest after travelling a distance of 50m. what is the force of friction between the stone and the ice.

Given,

Mass of stone = 1 kg

Initial velocity, u = 20 m/s

Final velocity, v = 0 (as stone comes to rest)

Distance covered, s = 50 m

Force of friction =?

We know that,

Now, we know that, force, F = mass x acceleration

Therefore, F = 1 kg X -4ms-2

Or, F = -4ms-2
Thus, force of friction acting upon stone = -4ms-2. Here negative sign shows that force is being applied in the opposite direction of the movement of the stone.
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Initial velocity, u = 20 m/s
Final velocity, v = 0 ( the stone stops )
Acceleration, a = ?   ( To be calculated )
And,    Distance travelled, s = 50 m
Now,    v² = u² + 2as
​So,     (0)² = (20)² + 2 x a x 50
                0  = 400 + 100 a
          100 a = -400
                 a = -400/100
                 a = -4 m/s²
Now,        Force, F = mass x acceleration
                           F = 1 x (-4) N
                          F = -4 Newtons
Thus, the force of friction between the stone and the ice is 4 newtons. The negative sign shows that this force opposes the motion.
 
Hope you Understand !

             

 

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