(3%) Problem 20: A horizontal force, F,-55 N, and a force, F2 = 10.8 N acting at
ID: 1771927 • Letter: #
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(3%) Problem 20: A horizontal force, F,-55 N, and a force, F2 = 10.8 N acting at an angle of to the horizontal, are applied to a block of mass 3.5 kg. The coefficient of kinetic friction between the block and the surface is k = 0.2. The block is moving to the right. Randomized Variables F1=55 N F2-10.8N m-35 kg ©theexpertta.com 50% Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if = 30° Grade Summary Deductions Potential 0% 100% tan() | | ( cotan0 asinO atan0 acotansinhO coshO tanh0 cotanhO 456 1 2 3 Submissions Attempts remaining:5 (390 per attempt) detailed view 0 END Degrees Radians 4o BACKSPACE DEL CLEAR Submit I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: % deduction per feedback. 50% Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if = 30%Explanation / Answer
a) Normal force FN = F + Ff = m g + (F2 * sin theta)
= (3.5 * 9.8) + (10.8 * sin 30)
FN = 39.7 N
b) F = F1 - F2 cos theta - mu FN
= 55 - (10.8 * cos 30) - (0.2 * 39.7) = 37.71 N
a = F / m = 37.71 / 3.5
acceleration = 10.77 m/s2
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