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 = 0.630 and μ_k = 0.530 and (b) μ_s = 0.430 and μ_k = 0.320?

(a)

Number

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Units

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(b)

Number

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Units

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Answer 1

A.)

Max frictional resisting force = μs . N

N = mg - (0.5 . m . g . sin 20 ) the bit in brackets is the upward component of the pulling force.

so max friction ( always the static friction) = 0.63.m.g( 1 - 0.5 . sin20)

= 0.63 . 0.829 . mg = 0.522 mg

But the horizontal pulling force is 0.5 . mg . cos 20 = 0.47 mg

Didn't move - the max possible static friction was bigger than the pulling force. so, acceleration = 0.

B.) In this case , N would be the same as part A.

max friction ( always the static friction) = 0.43.m.g( 1 - 0.5 . sin20) = 0.43 . 0.829 . mg = 0.356 mg

the pulling force is bigger than the maximum friction so the block will start to move.

Once the block is moving the coeff of friction that applies is the kinetic one. So the total accelerating force on the block is Pulling force - Friction

=( 0.5 . mg . cos 20 ) - 0.32( mg - (0.5 . m . g . sin 20 )

= 0.470 mg - 0.265 mg

= 0.204 mg

Now,

F = m . a

0.204 mg = m . a

So, a = 0.204g = 2.04 m/s²

^{Use; g = 10 m/s^2.}