Question

A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
-16N

Answer 1

Answer:

-4N (4N force in the opposite direction)

Explanation:

s = 50m

u =  20 m/s

v = 0 m/s

m = 1kg

By the motion formula $2as = v^{2} - u^{2}$, we can say:

2 × a × 50 = 0² - 20²

100a = -400

a = -4 m/s (Deceleration)

Force exerted by friction:

$F = ma$

F = 1  × -4

F = -4 Newton, i.e 4 Newton force in the opposite direction.

I hope this helps!!

Regards,

Lilac584