Question

b. A stone of 1kg is thrown with a velocity of 20ms across the frozen surface of a
lake and comes to rest after travelling a distance of som. What is the force of
friction between the stone and the ce?​

Answer 1

Answer:

Before we proceed we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force is acting in leftward while the stone in moving rightward direction. One thing we have to fix in our mind that frictional force will always act opposite to the direction of motion of the body.

Now on resolving the forces,

f−ma or f=ma                         ------(1)

also form the equation of motion v  

2

=u  

2

+2as, here final velocity is zero,so v=0

u  

2

=−2as

a=  

2s

−u  

2

 

​

=−  

2×50

20  

2

 

​

=−4  

sec  

2

 

m

​

 

on substituting the given data in equation(1)

f=1×(−4)=−4N