Question

A stone of 2kg is thrown with a velocity of 20m/s across the frozen surface of a lake and comes to rest after travelling the distance of 50 m. What is the force of friction exerted by the ice?

Answer 1

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Let ‘v’ be the speed with which the stone hits the ground. It falls through a height 40 m.

Using, v2 = u2 + 2as

=> v2 = 0 + 2 × 9.8 × 40

=> v = 28 m/s

Now, the stone penetrates 1 m into sand and comes to rest. Its acceleration is say ‘a’ and time to stop is say ‘t’.

Using, v2 = u2 + 2as

=> 0 = 282 + 2 × a × 1

=> a = -392 m/s2

Since, v = u + at

=> 0 = 28 – 392t

=> t = 0.07 s

Answer 2

Given:

• A stone of 2kg is thrown with a velocity of 20m/s.
• It comes to rest after traveling 50m.

To Find:

• The force exerted by the friction on ice.

Answer:

→ Mass if stone = 2kg.

→ Velocity ( initial) = 20m/s.

→ Velocity ( Final ) = 0m/s.

→ Distance covered = 50m

Now using third equation of motion ,

★ 2as = v ² - u² ★

=> 2 × a × 50m = (0)² - (20m/s)².

=> 100a m = - 400m²/s².

=> a = -400/100 m/s².

=> a = -4 m/s².

Hence the acclⁿ is -4 m/s².

Now we know formula of force as ,

★ Force = mass × acclⁿ ★

=> F = 2kg × -4m/s²

=> F = -8 N .

Hence the required answer is 8N .