Question

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

Answer 1

Answer:

(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.