Question

a stone of 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 50 km. what is the force of friction between the stone and the ice?
my doubt is....when we find the force...we get it as -4N.
frictional force acts in opposite direction...so will the frictional force be equal to -4 or +4?

Answer 1

Given :-

▪ A stone of 1 kg is thrown with a velocity of u = 20 m/s accross a frozen surface of a lake and comes to rest (i.e., final velocity, v = 0 m/s )

To Find :-

▪ Force of friction between the stone and the ice.

Solution :-

Given us the distance travelled, initial velocity and final velocity, let us find the acceleration using the third equation of motion.

We have,

• u = 20 m/s
• s = 50 m [ Mistake in Question ]
• v = 0 m/s

⇒ 2as = v² - u²

⇒ 2 × a × 50 = 0 - (20)²

⇒ 100a = -400

⇒ a = -4 m/s²

So, The acceleration of stone is -4 m/s² which is provided by the surface of the lake.

Now, We know

⇒ Force = Mass × acceleration

⇒ F = 1 × -4

⇒ F = -4 N

Hence, The force of friction between the stone and the surface of the lake is -4 newton.