Question

(5%) Problem 2: Ahorizontal force, Fi = 75 N, and a force, F2 = 19.9 N acting at an angle of 0 to the horizontal, are applied to a block of mass m =3.9 kg. The coefficient of kinetic friction between the block and the surface is u = 0.2.The block is moving to the right F. Randomized Variables F F1 75 N F2 = 19.9 N m 3.9 kg Otheexpertta.com 50% Part (a) Solve mumerically for the magnitude of the normal force, Fv in Newtons, that acts on the block if 0 = 30° Grade Summary Deductions Potential 0% 100% Submissions sin) tan ( 7 HOME Attempts remaining: 5 (5% per attempt) detailed view cotan asin acos() A 4 6 sinh atan acotan) 1 2 3 cosh tanh ODegrees cotanh END + . Radians VO BACKSPACE DEL CLEAR Submit Feedback I give up! Hint Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback 50% Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if 0 30°.

Answer 1

normal force.

N = mg + F2 sin x

N = 3.9* 9.8 + 19.9* sin 30

N = 48.17 N

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b)

net force along horizontal

ma = F1 - F2 cos x - uN

3.9 a = 75 - 19.9* cos 30 - 0.2* 48.17

a = 12.34 m/s^2

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