Question

A block is placed on a wet slippery floor. The block is 15 kg. When it is pulled through a string and spring balance, it shows force equal to 3 N. Find the coefficient of friction. (Fs = u mg).​

Answer 1

Answer:

GIVEN-

Mass of block = 15kg

Force = 3N

TO FIND-

The coefficient of FRICTION (u) ?

FORMULA USED -

F = u × N

(N = normal force)

Solution-

F = uN

F = u mg

u = F / mg

u = 3 / 15 × 10

u = 1 /50

u = 0.2

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Answer 2

Given: A block is placed on a wet slippery floor. The block is 15 kg. When it is pulled through a string and spring balance, it shows force equal to 3 N.

To find: The coefficient of friction.

Solution:

• Friction is the force that comes into play when one body slides or rolls over the surface of another body and acts in a direction tangential to the surfaces in contact and opposite to the direction of motion of the body.
• Frictional force is given by the formula,

        $F = \mu mg$

• Here, F is the frictional force, μ is the coefficient of friction, m is the mass of the body and g is the acceleration due to gravity.

        $3 N = \mu * 15 kg * 10 ms^{-2}$

         $\mu = 0.02$

Therefore, the coefficient of friction is 0.02.