Ablock weighing 8.5 n requires a force of 2 nto push it along at constant velocity. what is the coefficient of friction for thesurface?

The coefficient of friction of the surface μ = 0.235

Explanation:

Values of the given variables:

Weight of block = 8.5 N

Pushing force F = 2 N

velocity = constant

Solution:

Newton's Second Law of motion states that acceleration of an object depends on the mass of the object and the force applied

F = m a

Since v =constant, a = 0. Thus

Net force Fnet = 0.

Now F_net is the sum of the pushing force and the frictional force along the horizontal, which opposes the motion:

Fnet = Fpushing - Ffrictional_force

Given that Fnet = 0, Fpushing = Ffrictional_force. and Ffrictional_force = 2 N

Equation of friction is given by

Ffrictional_force = μ*W

2 = μ*8.5

μ = 2/8.5 = 0.235

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acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.