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?

Answer 1

Answer :-

-4N

Explanation :-

Given :

• Mass of the stone => 1kg

• Initial velocity of the stone => 20m/s

• Final velocity of the stone => 0

To Find :

• Force = ?

Solution :

We know,

$\boxed{\sf{f = ma}}$

here,

f is the force applied,

m is the mass and,

a is the acceleration.

To find force,we must be knowing acceleration,so let’s find out it’s acceleration first.

According to the third equation of motion,

$\boxed{\sf{v^2=u^2+2as}}$

here,

v is the final velocity,

u is the initial velocity,

a is the acceleration and,

s is the distance travelled.

So,

$\sf{}0=(20)^2+2\times a\times 50$

$\sf{}:\implies 0=400+100a$

$\sf{}:\implies -400=100a$

$\sf{}:\implies -\dfrac{400}{100}=a$

$\sf{}\therefore a =-4m/s^2$

The negative sign indicates that acceleration is acting against the motion of the stone.

Force,

$\sf{}:\implies 1\times (-4)$

$\sf{}\therefore -4N$

Hence, the force of friction between the stone and the ice is −4 N.