Question

A stone of mass 0.5 kg is thrown with a velocity of 10 m/s across the frozen surface of a lake and it comes to rest after travelling a distance of 100 m. What is the magnitude of force of friction between the stone and ice?

Answer 1

Answer:

Given:

• Mass = 0.5 kg
• Initial Velocity = 10 m/s
• Final Velocity = 0 m/s
• Distance Travelled = 100 m

To Find:

• Magnitude of Frictional Force

Solution:

Force = Mass × Acceleration

Therefore we need to find acceleration. Substituting in Third Equation of Motion we get,

⇒ 0² - 10² = 2 ( a ) ( s )

⇒ -100 = 2 ( a ) ( 100 )

⇒ 2a = -1

⇒ a = -1/2 = -0.5 m/s²  [ - sign indicates retardation ]

Force with which the object was travelling is:

⇒ Force = 0.5 × -0.5 = -0.25 N

Therefore the opposing force is indicated by the negative sign. Thus it is friction. Hence magnitude of friction is 0.25 N.

Hope it helped !!

Answer 2

Explanation:-

Given :-

mass of stone = 0.5 kg

initial velocity = 10 m/s

Final velocity = 0m/s

Distance travelled = 100 m

To find :-

The magnitude of force of friction.

Solution:-

Let us find the acceleration of the particle.

• By using equation of motion.

$\boxed{2as = v^2 - u^2}$

• Put the given values.

→$2 \times a \times 100 = (0)^2 -(10)^2$

→$200a = -100$

→$a = \dfrac{-100}{200}$

→$a = -0.5 m/s^2$

acceleration of stone will be

-0.5 m/s².

The magnitude of force of friction which oppose the stone is given by :-

→$\boxed{ F_{f} = ma }$

• By using Newton 2nd law.

→$F_{f} = 0.5 \times -0.5$

→$F_{f}= -0.25 N$

Hence,

The magnitude of force of friction is 0.25 N.