sledge of mass 80 kg moves over the frozen surface of a lake with a velocity of 30m/s and comes to rest in 5s. find frictional force between the lake and sledge

Answer 1

Average acceleration
=(0-30) m/s ÷ 5 s
= -6 m/s²

Force
= ma = 80 kg * (-6 m/s²)
= -480 N (resistive force)

Answer 2

F=ma
F=80×-30/5
F=80×6=-480N
Declaration=change in velocity
/time -30/5=-6mssquare