Question

a stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m . what is the force friction between the stone and the ice?

Answer 1

Answer:

Mass = 1 kg

Initial Velocity = 20 m/s

Distance traveled = 50 m.

Since the object comes to rest at the end, the final velocity is 0 m/s.

Frictional Force = Mass * Acceleration

V² - U² = 2as

=> 0² - 20² = 2 ( 50 ) ( s )

=> -400 = 100 a

=> a = - 400 / 100 = -4 m/s²

=> Force = 1 kg * -4 m/s² = -4 kg.m/s² or -4 N

Hence the frictional force is -4 N.

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\green{Answer:-}$

Here, m= 1kg, u= 20 m s^-1, v= 0, s= 50m

As, v² - u² = 2as

Therefore, 0² - 20² = 2a × 50

or a= - 400/100 = -4 ms^-2

Now, Force of friction, F= ma= 1 × (-4) = -4 N.

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