Question

Stone size 5kg is thrown with a velocity of 16 m/s across the frozen of a lake and come to rest after traveling a distance 50 m . What is the force of friction between the stone and ice

Answer 1

Answer:

• Friction force between stone and ice = -12.8 N

[ Negative sign shows that force is acting opposite to the direction of motion of stone ]

Explanation:

Given

• Mass of stone, m = 5 kg
• Initial velocity of stone with which it is thrown, u = 16 m/s
• Distance covered by stone, s = 50 m
• Final velocity of stone, v = 0

To find

• Force of friction between the stone and ice. F =?

Formula required

• Third equation of motion

        2 a s = v² - u²

• Newton's second law

       F = m a

[ Where a is acceleration, s is distance covered, v is final velocity, u is initial velocity, F is force applied, m is mass ]

Solution

Let, acceleration of stone while thrown across the frozen lake be a

Then, using third equation of motion

→ 2 a s = v² - u²

→ 2 a ( 50 ) = ( 0 )² - ( 16 )²

→ 100 a = -256

→ a = -2.56 m/s²

So, the acceleration of stone due to the friction of frozen lake will be -2.56 m/s².

Now, By newton's second law

→ F = m a

→ F = ( 5 ) · ( -2.56 )

→ F = -12.8  N

Therefore,

• A friction force of magnitude 12.8 Newton would be acting between stone and ice in the direction opposite to the motion of stone.

Answer 2

Answer:

Given:

Initial velocity of the stone, u= 16 m/s

Final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

We know the third equation of motion

v²=u²+2as

Substituting the known values in the above equation we get,

0² = (16)²+2(a)(50)

-256 = 100a

a = -256/100 = -2.56m/s² (retardation)

We know that,

F = m×a

Substituting above obtained value of a = -2.56 in F = m × a.

We get,

F = 5×(-2.56)

F = -12.8N