Answer:
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
