Question

2. A coin of mass 20 g is pushed on a table. The coin starts moving at a speed of 25 cm/s and stops in 5 seconds. Find the force of friction exerted by the table on the coin.​

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Mass of the coin = 20 g

✭ Initial Velocity (u) = 25 cm/s

✭ Final Velocity (v) = 0 cm/s

✭ Time (t) = 5 sec

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ Frictional force?

$\displaystyle\large\underline{\sf\gray{Solution}}$

So here the mass of the body will be,

✪ 20/1000 = 0.02 kg

Initial velocity will be,

✪ 25/100 = 0.25 m/s

And the final velocity will be,

✪ 0 m/s

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

We shall first find the acceleration of the body with the help of the first Equation of motion, that is,

$\displaystyle\underline{\boxed{\sf v = u+at}}$

• v = 0 m/s
• u = 0.25 m/s
• t = 5 sec

Substituting the values,

➝ $\displaystyle\sf v = u+at$

➝ $\displaystyle\sf 0 = 0.25+a\times 5$

➝ $\displaystyle\sf -0.25 = 5a$

➝ $\displaystyle\sf \dfrac{-0.25}{5} = a$

➝ $\displaystyle\sf \orange{Acceleration = -0.05 \ m/s^2}$

Now the frictional force will be given by,

$\displaystyle\underline{\boxed{\sf F = ma}}$

• m = 0.02 kg
• a = -0.05

Substituting the values,

➳ $\displaystyle\sf F = ma$

➳ $\displaystyle\sf F = 0.02\times -0.05$

➳ $\displaystyle\sf F = -0.001$

➳ $\displaystyle\sf \pink{F = 0.001 \ N}$

$\displaystyle\sf \therefore\:\underline{\sf Frictional \ Force \ is \ 0.001 \ N}$

Note : So here we have ignored the negative sign because friction itself is an opposing force

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