Question

A block of mass 2 kg placed on an inclined plane just slides down when its angle inclination is 45° calculate the
(i) coefficient of friction between the surfaces of incline plane in block
(ii)acceleration of the block when angle of inclination of the plane is 30 degree

Answer 1

1)coefficient of friction is 1

2) acceleration is (√3-1)g/2 m/s²

See attachment for solution :

Answer 2

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♦ As given that

• Mass of block = 2 kg

• Inclination of plane = 45°

♦ Then to find out

(i) Coefficient of friction

♦ As the block just starts moving we will take

$F_s = F_{g}$

>> Where

$F_s = \textbf{Limiting friction}$

$F_g = \textbf{Force of attraction of gravity}$

♦ Now as

$F_s = \mu N$

$\implies F_s = \mu Mass \times g \times Cos \:45$

Also

$F_g = Mass \times g \times Cos \:45$

Then by comparing we get

$F_s = \mu F_g$

and as $F_s = F_{g}$

$\implies \mu = \dfrac{F_s}{F_g}$

$\implies \mu = 1$

(ii)

First we will find out Limiting friction

As

$F_s = \mu N$

$\implies F_s = 1 \times 2 \times 9.8 \times sin 30$

$\implies F_s = 2 \times 9.8 \times \dfrac{1}{2}$

$\implies F_s = 9.8$

Now As $F_g = Mass \times g \times Cos 30$

$\implies F_g = 2 \times 9.8 \times \dfrac{\sqrt{3}}{2}$

$\implies F_g = 9.8\sqrt{3}$

♦ So total Force working

$= F_s - F_g \\\\ = 9.8 - 9.8\sqrt{3} \\\\ = 9.8(1-\sqrt{3})$

♦Now as acceleration

$= \dfrac{Total\: Force}{Mass}$

$= \dfrac{9.8(1-\sqrt{3})}{2}$

♦ So , it will have a acceleration of $\dfrac{9.8(1-\sqrt{3}}{2}) m/s^2$ or in the direction of gravitation force of earth or opposite of the direction of friction .