Question

A stone at mass 500g is thrown with the a velocity of 20 m/s across a frozen surface of lake. It comes to rest at by travelling a distance of 0.1 km. Calculate force of friction between the stone and frozen surface of lake. ​

Answer 1

Answer:

The stone is sliding on the ice and comes to rest that means there is retardation....

so, by the third equation of motion

v=0

u=20 m/s

s=0.1 km = 100 m

m= 500g = 0.5 kg

v^2=u^2-2as

=> 0=20-2×a×100

=> 0=20-200a

=> 20=200a

=> a= 0.1 ms^-2

Force of friction= m×a

=0.5×0.1

= 0.05 N

Explanation:

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