Question

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◾A stone of 1kg is thrown with a velocity of 20 m/s . Across a frozen surface of a lake and comes to rest after traveling a distance of 50 m . what is the force of friction .? ​

Answer 1

mass =1kg

velocity=20m/s

force-friction=0

force=ma

1×(20-0)

20N

friction is equal to 20

Answer 2

Answer:

- 4 N

Explanation:

Given :

Initial velocity (  u ) = 20 m /sec

Final velocity ( v ) = 0 m / sec

Distance ( s ) = 50 m .

Mass ( m ) = 1 kg

In order to find force we have to find acceleration first :

From third equation of motion :

We have :

v² = u² + 2 a s

Putting values here :

0 = 400 + 2 × 50 × a

100 a = - 400

a = - 4 m / sec²

Now force = mass × acceleration

F = 1 × - 4 N

F = - 4 N .

Negative sign the show opposite direction only :  [ Force is also vector quantity ]

Hence we get answer .