Question

A stone of size 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 500 m. What is the force of friction between the stone and the ice​

Answer 1

QUESTION:-

A stone of size 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 500 m. What is the force of friction between the stone and the ice​

EXPLANATION:-

• Initial velocity(u)=20 m/s
• Final velocity(v)=0 m/s [AS THE BALL STOPS]
• Mass(m)=1 kg
• Distance(s)=500 m
• Force=?
• Acceleration(a)=?

We know that:-

F=ma

Where,

F=force

m=mass

a=acceleration

In the question ,

mass(m)=1 kg

acceleration(a)=?

So know let's calculate the acc. ,using 3rd equation of motion:-

v²=u²+2as

putting value,

(0)²=(20)²+2a(500)

0=400+1000a

-400=1000a

a=-400/1000

a=-2/5

a=-0.4 m/s²

So the acc. is -0.4 m/s²

Now we can find force of friction.

F=ma

F=1×-0.4

F=-0.4 N

SO the force b/s is -0.4 N

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