Question

please send me right answer with the explanation if you can't do it don't answer​

Answer 1

Question :

A stone of 1 kg is thrown with a velocity of 20 m/s 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 ?

Given :

• Initial velocity (u) = 20 m/s
• Final velocity (v) = 0 m/s
• Acceleration (a) = ?
• Distance travelled (s) = 50 m

To find :

• Force of friction between the stone and the ice.

So,according to the question at first we have to find acceleration

$\implies\sf{ {v}^{2} = {u}^{2} + 2as }$

$\implies \sf{ {(0)}^{2} = {(20)}^{2} + 2 \times a \times 50 }$

$\implies\sf{0 = 400 + 100a}$

$\implies \sf{0 - 400 = 100a}$

$\implies \sf{ - 400 = 100a}$

$\implies\sf{a = \frac{ - 4 \cancel{00}}{1 \cancel{00}} }$

$\implies \sf{ - 4 \: m {s}^{ - 1} } \green \bigstar$

Now we will calculate the force,

$\sf \implies{F = ma}$

• F stands for Force
• a stands for Acceleration
• m stands for Mass

$\sf \implies{ 1 \times ( - 4)}$

$\sf \implies{ - 4 \: N} \green \bigstar$

So,the force is -4 N it means that thus force opposes the motion of stone..