Question

Q1.A mass of 3 kg rests on horizontal plane. The plane is gradually inclined until at an angle of 37º to horizontal, the mass just begins to slide down. The coefficient of static friction between block and surface is:

Answer 1

We know that,

Mass m=4kg

angle θ=15

0

Now, the normal force is

N=mg

N=4×10

N=40N

Now, frictional force always opposes direction of motion

So,

F

f

=mgcosθ

μN=4×10×cos15

0

μ×40=40×0.96

μ=0.96

Hence, the friction is 0.96

it's khushi ❤️

Answer 2

Step-by-step explanation:

Given:

A mass of 3 kg rests on horizontal plane. The plane is gradually inclined until at an angle of 37º to horizontal, the mass just begins to slide down.

To find:

Coefficient friction between block and surface?

Calculation:

Just before the block starts to move, the block was in equilibrium. So, we can say:

$\rm \: mg \sin( \theta) = f$

$\rm \implies \: mg \sin( \theta) = \mu N$

$\rm \implies \: mg \sin( \theta) = \mu \{mg \cos( \theta) \}$

$\rm \implies \: \sin( \theta) = \mu \cos( \theta)</p><p>$

$\rm \implies \: \mu = \tan( \theta)$

$Putting \: \theta = 37°.$

$\rm \implies \: \mu = \tan( {37}^{ \circ}$

$\rm \implies \: \mu = \dfrac{3}{4}$

$\rm \implies \: \mu = 0.75$

So, coefficient of friction is 0.75.