the figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20∘. What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a) μs=0.610 and μk=0.500 and (b) μs=0.400 and μk=0.300

(a) 1.054 m/s²

(b) 1.404 m/s²

Explanation:

0.5·m·g·cos (θ) - μs·m·g· (1 - sin (θ)) - μk·m·g· (1 - sin (θ)) = m·a

Which gives;

0.5·g·cos (θ) - μ·g· (1 - sin (θ) = a

Where:

m = Mass of the of the block

μ = Coefficient of friction

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block

θ = Angle of elevation of the block = 20°

Therefore;

0.5*9.81·cos (20°) - μs*9.81 * (1 - sin (20°) - μk*9.81 * (1 - sin (20°) = a

(a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;

0.5*9.81·cos (20°) - 0.610*9.81 * (1 - sin (20°) - 0.500*9.81 * (1 - sin (20°) = 1.054 m/s²

(b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;

0.5*9.81·cos (20°) - 0.400*9.81 * (1 - sin (20°) - 0.300*9.81 * (1 - sin (20°) = 1.404 m/s².