Question

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Answer 1

Question:

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

Solution:

$\star$ Given

→ Mass (m) = 1 kg

→ Initial Velocity (u) = 20 m/s

→ Distance travelled (s) = 50 m

As the object comes to rest at the end,

Therefore, final velocity is 0 m/s.

Here, the Force (F) is frictional Force

Force = Mass * Acceleration

⇒ Acceleration = ?

The formula to be used is,

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

⇒ 0² - 20² = 2 ( 50 ) ( s )

⇒ - 400 = 100 a

⇒ a = - 400 / 100

a = - 4 m/s²

Therefore, Frictional Force = Mass * Acceleration

⇒ Force = 1 kg * - 4 m/s²

→  - 4 kgm/s² or - 4 N

$\boxed{Hence \;\;the\;\; frictional\;\; force\;\; is - 4 N.}$

Answer 2

Explanation:

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