Question

a stone of mass 500g is thrown with a velocity of 20 m/s across the frozen surface of a lake comes to rest after travelling a distance of 0.1km. calculate force of friction between the stone and the frozen surface of lake.

Answer 1

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

Answer 2

Answer:

The force of friction between the stone and the frozen lake is equal to 1N.

Explanation:

We have given,

The mass of the stone, m = 500g = 0.5kg

The initial velocity of the stone, u = 20m/s

The distance travelled by the stone before comes to rest:

S = 0.1 Km = 100m

From the third equation of motion:

$v^2-u^2=2aS$

$(0)^2-(20)^2 = 2a(100)$

$200a= -400$

$a= -2ms^{-2}$

The negative sign shows the retardation in the motion of stone.

From the Newton's 2nd law:

F = ma

F = (0.5kg)×(-2ms^{-2})

$F =-1N$

Therefore, the force of friction is 1N. The negative sign shows that the force of friction is in opposite direction to direction of motion of the stone.