Question

A stone of size 1 kg is thrown with a velocity of 20 metre per sec across frozen surface of lake and it comes to rest after travelling a distance of 50 m what is force of friction between them

Answer 1

Answer:

1kg through the velocity of 20 metre per second across the five minutes like and V what is the force of between them is

Answer 2

$\huge\underline{\overline{\mid{\bold{\orange{Given-}}\mid}}}$

• Initial velocity of the stone, u= 20 m/s
• Final velocity of the stone, v= 0
• Distance covered by the stone, s= 50 m

$\huge\underline{\overline{\mid{\bold{\blue{Formula-}}\mid}}}$

• $\boxed{v^2=u^2+2as}$

• $\boxed{F=ma}$

$\huge\underline{\overline{\mid{\bold{\red{Solution-}}\mid}}}$

Putting all values in formula we get,

$0^2 = (20)^2+2(a)(50)$

$-400 = 100a => -4m/s^2$

we know the formula of force ,

F=ma

putting value in this formula we get,

F = 1×(-4) = -4N

hence, force of friction between them = -4N