A 75kg encounters 10n of friction and slides down a hall with an acceleration of 3.60 find the coefficients of kinetic between the box and floor

Explanation:

Given,

Mass of block = 75kg

Force of friction=10N

Acceleration of box = 3.60m/s^2

Acceleration due to gravity = 9.8m/s^2

Let tetha represent the angle of inclination.

Wherefore, we have mgsinθ - force of friction = ma 1

Substitute the values into equation 1

75×9.8×sinθ -10 = 75×3.60

735sinθ = 10+270

735sinθ = 280

Divide both sides by 735

Sinθ = 280/735

Sinθ = 0.3809

θ = sin^-1 0.3809

θ = 22.389°

We can now solve for the coefficient of friction by considering the formula:

Fs = us×R

Where R = mgcosθ

Fs = 10

10 = Us×75×9.8×cos22.389

10 = Us× 735×0.9246

10 = Us × 679.595

Divide both sides by 679.595

10/679.595 = Us

Us = 0.01471

Hence the coefficient of friction is 0.01471

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