HW 10 - Part A Begin Date: 3/15/2018 12:01:00 AM Due Date: 3/24/2018 11:59:00 PM
ID: 2029808 • Letter: H
Question
HW 10 - Part A Begin Date: 3/15/2018 12:01:00 AM Due Date: 3/24/2018 11:59:00 PM End Date: 3/31/2018 11:59:00 PM (10%) Problem 1: A block of mass m is initially at rest at the top of an inclined plane, which has a height of 5.6 m and makes an angle of e 29 with respect to the horizontal. After being released, it is observed to be traveling atv0.15 m/s a distance d after the end of the inclined plane as shown. The coefficient of kinetic friction between the block and the plane is0.1, and the coefficient of friction on the horizontal surface is -0.2 initial Final Otheexpertta.com Find the distance d. in meters. Grade Summa Deductions Potential 496 9696 7 8 9 tan() | acosO cosO Submissions Attempts remaining: 10 (0per attempt) detailed view Sin cotan asin0 atanacotan sinhO coshO anh cotanhO Degrees Radians Submit Hint I give up! Hints: 1 for a 4% deduction. Hints remaining: 1 Feedback: 500-deduction per feedback. Use conservation of energy in the presence of dissipative forces. Any energy lost here is due to friction, in the form of work.Explanation / Answer
l = h / sin theta = 5.6 / sin 29 = 11.55 m
acceleration a = g sin theta - mu g cos theta
= (9.8 * sin 29) - (0.1 * 9.8 * cos 29) = 3.89 m/s2
speed of box v = sqrt [2 a l]
v = sqrt [2 * 3.89 * 11.55] = 9.48 m/s
m a = - mu m g
a = - mu g
v2 = v02 + 2 a d
d = -v02 / 2 a = v02 / 2 (mu * g)
d = 9.482 / (2 * 0.2 * 9.8)
distance d = 22.9 m
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