Question

A force of 120 newton is required to pull a body of
mass 1x10^3 kg over the surface of the ice then
find the co-efficient of friction (g = 7.6 ms-)​

Answer 1

✭ $\mathcal{\huge{\underline{\underline{\red{Question:-}}}}}$

✒ A force of 120 newton is required to pull a body of mass 1x10³ kg over the surface of the ice then. Find the co-efficient of friction? (g = 7.6 ms-¹)

✭ Answer :-

⬛ The co-efficient of friction is 0.015 .

✭ Solution :-

The block is placed on the horizontal plane, so

$\mathcal{\huge{\fbox{\fbox{\blue{mg\:=\:N}}}}}$

Also maximum friction force is μ.N, which is

120 N.

Friction force will be equal to force applied until it reaches its maximum value. After that limiting friction will act and body will start moving.

So, μ.N = 120 N

▶μ.×1×10³×7.6=120

▶μ.×1000×7.6=120

▶μ × 7600 = 120

▶μ = $\dfrac{120}{7600}$

▶μ = $\cancel{\dfrac{120}{7600}}$

✏ μ = 0.015

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