The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F = 0.550mg is then applied at upward angle θ = 23°.
(a) What is the magnitude of the acceleration of the block
across the floor if the friction coefficients are
μs = 0.590 and
μk = 0.495?
m/s2
(b) What is the magnitude of the acceleration of the block across
the floor if the friction coefficients are
μs = 0.395 and
μk = 0.295?
m/s2
max fricitonal force
f = u ( mg - F sin x)
f = 0.59* ( mg - 0.55 mg sin 23)
f = 0.4632 mg
horizontal force applied
Fh = 0.55 cos 23 = 0.5063 mg
since static fricitonal force is lesser than applied horizontal force so kinetic fricitonal force acts on the object
m a = Fh - 0.495* ( mg - 0.55 mg sin 23)
a = 1.153 m/s^2
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b)
max static frictional force
f = 0.395* ( mg - 0.55 mg sin 23)
f = 0.31 mg
since max fricitonal force is less than applied horizontal force so object can't move from its position,hence,
a = 0 m/s^2
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