Question

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

Answer 1

$\huge\mathcal\purple{Here \:is \:the\: answer}$

Refer to the attachment !!!!

Answer 2

$\huge\underline\mathfrak{Answer:}$

Initial velocity (u) = 20ms-¹

Final velocity (v) = 0

Distance travelled = 50m

Acceleration (a) = ?

$\sf{\underline{\underline{By\:using\:3rd\:equation\:of\:motion:-}}}$

$\implies$$\sf\frac{v^2-u^2}{2s}=a$

$\implies$$\sf\frac{(0)^2-(20)^2}{2\times50}=a$

$\implies$$\sf\frac{4\cancel00}{1\cancel00}=a$

$\bold{\large{\boxed{\sf{\pink{a=-4ms-^2}}}}}$

Force (F) = m$\times$a

= 1$\times$4

= -4 N

Force of friction = 4N (friction always acts in opposite direction of force.)