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 (a) μ_s = 0.610 and μ_k = 0.510 and (b) μ_s = 0.430 and μ_k = 0.310?

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 (a) As-0.610 and μk-0.510 and (b) μ.-0.430 and Pk- 0.310? (a) Number Units (b) Number Units

Answer 1

normal force

N = mg - F sin x

applied force alomg horizontal

Fx = 0.5 mg cos 20 = 0.47 mg

a)

max frictional force

Fs = u N = 0.61* (mg - 0.5 mg sin 20 ) = 0.83 mg

since static frictional force is less than applied force

so object won't move

a = 0 m/s^2

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b)

max frictional force

Fs = 0.43* ( mg - 0.5* mg sin 20) = 0.3565 mg

now, since Fs < Fx

so kinetic frictional force started to acts

ma = Fx - Fs

m a = mg ( 0.47 - 0.3565)

a = 0.1135 g

a = 1.112 m/s^2

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