Question

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

Answer 1

Givens:

mass of stone = 1kg

u = 20m/s. velocity = 0m/s.

s(distance travelled) = 50m

Solution:

use this 1st equation

v²=u²+2as

0² = (20)²+2(a)(50) =

-400 = 100a

a = -400/100 = -4m/s²

2nd equation

F = m×a

F = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)

Answer:

the force of friction between the stone and the ice is – 4 N.

I hope this helped!

Explanation:

Answer 2

50 m. What is the force of friction between the stone and the ice?

Initial velocity of the stone, u= 20 m/s

Final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

We know the third equation of motion

v²=u²+2as

Substituting the known values in the above equation we get,

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

We know that

F = m×a

Substituting above obtained value of $a=-4 in F= m x awe get,$

$F = 1×(-4) = -4N$

(Here the negative sign indicates the opposing force which is Friction)