# 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.

v=0,

u=20m/s,

s=50 m

m=1kg

using 3rd equation of motion

v2=u2+2as

02=202+2xax50

400+100a

-400=100a

-400/100=a

-4m/s2= a

f= mxa

f= 1x -4

f= -4 N  { "--"  minus sign shown the direction of force }

• 7

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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• -6
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 !

• 64
ALL TE 3 ANSWERS ARE CORRECT.
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Aa
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