Question

a stone of 1 kg is thrown with a velocity of 20 metre per second across the frozen surface of a lake and comes to the rest after travelling a distance of 50 metere what is the force of friction between the stone and icea stone of 1 kg is thrown with a velocity of 20 metre per second across the frozen surface of a lake and comes to the rest after travelling a distance 50 metre what is the force of friction between the stone and ice

Answer 1

Answer:

d= 50m

u=20

v=0

2as= v^2-u^2.

2×a×50=0-400

=100a= -400

a= -400÷100

a= -4m/s^2.

force of friction= 1×-4=-4N

Answer 2

Answer:

F=ma

initial velocity=20m/s

final velocity=0m/s

did. travelled=50m

v^2=u^-2as

we use -2as because acceleration is in opposite direction

0^2=20^2-2*50*a

-400=-100a

a=4m/s^2

F=ma

F=1*4

F=4N