Question

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

Explanation:

Answer

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

Answer 2

The coefficient of static friction between block and surface is 0.75

Explanation:

Given:

A mass of 3 kg rests on a horizontal plane

When the plane is inclined at 37º, the mass begins to slide down

To find out:

The coefficient of static friction between the block and surface

Solution:

Given

m = 3 kg

θ = 37º

When the block is on the verge of sliding at that moment

The component of the weight along the inclined plane = Frictional force

Thus,

$mg\sin\theta=\mu N$

$\implies mg\sin\theta=\mu mg\cos\theta$

$\implies \mu=\frac{\sin\theta}{\cos\theta}$

$\implies \mu=\tan\theta$

$\implies \mu=\tan37^\circ$

$\implies \mu=\frac{3}{4}$

$\implies \mu=0.75$

Hope this answer is helpful.

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