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.
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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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 !