Question

A stone of 2kg is thrown with a velocity of 20m/s across the frozen surface of a lake and comes to rest after travelling the distance of 50 m. What is the force of friction exerted by the ice​

Answer 1

Answer:

here is the answer for your question

Answer 2

Answer:

-8N

Explanation:

Given :

• Mass,m=> 2kg

• Initial velocity,u => 20m/s

• Final velocity,v => 0m/s

• Distance,s => 50m

To Find

• Force

Solution :

To find force first we have to find it’s accelaraction :-

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

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

$\rightarrow \sf{}0=400+2(a)(50)$

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

$\rightarrow \sf{}-\dfrac{400}{100}=a$

$\rightarrow \sf{}a=-4$

We know :-

$\sf{}F=ma$

Therefore,Force = >

$\to\sf{}F=2\times-4$

$\sf{}\therefore Force=-8N$

[Negative sign indicates that the force is applied in opposite direction]