Question

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

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Answer 1

Solution :-

• M = 1kg

• u = 20$\dfrac{m}{s}$

• s = 50m

• v = 0

• F = ?

• a= ?

⠀⠀⠀⠀⠀⠀⟹ $\boxed{ {v}^{2}\:-\: {u}^{2}\:= \:2as}$

⠀⠀⠀⠀⠀⠀⠀⠀⟹ ${(0)}^{2}\:-\:{(20)}^{2}\:=\:2a(50)$

⠀⠀⠀⠀⠀⠀⠀⠀⟹ (-400) = 100a

⠀⠀⠀⠀⠀⠀⠀⠀⟹ A = $\dfrac{400}{100}$

⠀⠀⠀⠀⠀⠀⟹-4$\dfrac{m}{s}^{2}$

Force of friction

⠀⠀⠀⠀⠀⠀⠀⠀= F = m × a

⠀⠀⠀⠀⠀⠀⠀⠀= 1 × -4$\dfrac{m}{s}^{2}$

⠀⠀⠀⠀⠀⠀⠀= -4N

hence, of force of friction between the stone and ice is -4n

Answer 2

$\huge\frak\red{Question}$

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

$\huge\frak\red{Solution★}$

Initial velocity of the stone, u= 20 m/s

Final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

We know the third equation of motion

v²=u²+2as

Substituting the known values in the above equation we get,

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

We know that

F = m×a

Substituting above obtained value of a=-4 in F= m x awe get,

F = 1×(-4) = -4N

(Here the negative sign indicates the opposing force which is Friction)