Question

A stone of 1 kg is thrown with a velocity of 40 m/s 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? *
A)16 N
B)-4 N
C)-16 N

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Answer 1

Answer:

C)-16 N

Explanation:

concepts used

force = mass* acceleration

equation of motion

$v^2 = u^2 + 2as$

where v is the final velocity

u is the initial velocity

and s is the distance moved

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Given

mass = 1 kg

initial velocity (u) = 40m/s

final velocity (v) = 0 as stones comes to rest

distance moved by stone (s) = 50m

using  $v^2 = u^2 + 2as$

$0 = 40^2 +2a*50\\=> -1600 = 100a\\=> a = -1600/100 = -16$

Thus, acceleration is -16 m/s^2

here acceleration is negative as force of friction is opposing the motion.

Force of friction = mass of stone * acceleration of stone

Force of friction = 1*-16 kgm/s^2 = -16N       ( kgm/s^2 = 1 N)

Thus, option c -16N is correct choice.