Question

A stone of 1 kg is thrown with a Velocity of 20 m/s acoross the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the Acceleration and force of Friction Between the stone & the ice.

Answer 1

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$\red{\underline{{ \bf Question }}}$

• A stone of 1 kg is thrown with a Velocity of 20 m/s acoross the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the Acceleration and force of Friction Between the stone & the ice.

$\underline {\underline{{\purple{ \sf Given }}}}$

• Mass of Stone (m) ➠ 1 Kg
• Initial Velocity (u) ➠ 20 m/s
• Final Velocity (v) ➠ 0 m/s
• Distance (s) ➠ 50 m

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• Acceleration
• Force of Friction

$\underline {\underline{{ \green{\sf Calculating \: Acceleration }}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf{v}^{2} - {u}^{2} = 2as }}}}$

Substituting Values

☞ $\blue{\sf (0)^{2} - (20)^{2} = 2 \times a \times 50 }$

☞ $\blue{\sf 0 - 400 = 100 \times a}$

☞ $\blue{ \sf -400 = 100a}$

☞ $\blue{ \sf 100a = -400}$

☞ $\blue{\sf a = \cancel{\dfrac{-400}{100}}}$

☞ $\blue{ \sf a = -4 }$

Therefore, acceleration is -4 m/s²

$\underline {\underline{{ \green{\sf Calculating \: Force \: of \: Friction }}}}$

Formula Used :- $\underline{\underline{\boxed{\pink{\sf Force(F) = Mass (m) \times Acceleration (a) }}}}$

Substituting Values

☞ $\blue{\sf F = 1 \times -4}$

☞ $\blue{\sf F = -4}$

Therefore, force of Friction is -4 Newton

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Answer 2

Answer:

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)

Explanation: