A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be

- 20

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- 0.2

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Solution

The correct option is C $- 0.2$ N
Let the friction force be $f$ $N$ .
Acceleration in the stone $= \frac{0 - 2}{10} = - 0.2 m s^{- 1}$
By Newton's Second Law:
$\Rightarrow f = m a$
$\Rightarrow f = 1 \times (- 0.2)$
$\Rightarrow f = - 0.2 N$

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A stone of mass 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake. It comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

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A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be

The correct option is C −0.2 NLet the friction force be f N.Acceleration in the stone =0−210=−0.2 ms−1By Newton's Second Law:⇒f=ma⇒f=1×(−0.2)⇒f=−0.2 N

The correct option is C $- 0.2$ N
Let the friction force be $f$ $N$ .
Acceleration in the stone $= \frac{0 - 2}{10} = - 0.2 m s^{- 1}$
By Newton's Second Law:
$\Rightarrow f = m a$
$\Rightarrow f = 1 \times (- 0.2)$
$\Rightarrow f = - 0.2 N$