The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F = 0.540·mg is then applied at upward angle θ = 20°.
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(a) What is the magnitude of the acceleration of the block across the floor if (a) μs = 0.600 and μk = 0.500?
(b) What is the magnitude of the acceleration of the block across the floor if μs = 0.400 and μk = 0.300?
would like steps and solutions. Thanks!
a) on block in vertical direction,
N + 0.540mg sin20 -mg =0
N = 0.815mg
In horizontal,
Fcos20 - friction = ma
Fcos20 = 0.540mgcos20 =0.507mg
maximumstatic friction = us. N = 0.6 x 0.815mg = 0.489mg
so maximum friction value is less than horizontal force .
block will start to move and kinetic friction will work.
kinetic friction = uk N = 0.500 x 0.815mg = 0.4075mg
0.507mg - 0.4075mg = ma
a = 0.976 m/s^2
b) now value of kinetic friction = 0.300 x 0.815mg = 0.2445mg
0.507mg - 0.2445mg = ma
a = 2.58 m/s^2
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