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| The Expert TA I Human-like Grading. Automated! Home I Student: spw0012& auburn.edu My Account Log Out Class Management I Help HW 4 Begin Date: 9/19/2018 12:01:00 AM - Due Date: 9127/2018 11:59:00 PM End Date: 12/16/2018 11:59:00 PM (8%) Problem 4: A block of mass m is initially at rest at the top of an inclined plane which has a height of 6.4 m and makes an angle of -29" with respect to the horizontal. After being relcased, it is observed to be traveling at v 0.35 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 is ,-0.1 , and the coefficient of friction on the horizontal surface is per-02. initial Final Assignment Status Click here for detailed view ©theexpertta.com Problem Status 1 Completed | | 50% Part (a) what is the speed of the block, in meters per second, just after it leaves the inclined plane? Grade Summary bottom 3 Completed 0% 100% cos0 cotan asin acosO atan acotan sinh cosh0 tan cotanh) Degrees Radians sinO tanO T Submissions Attempts remaining: 8 (2% per attempt) detailed view 1 23 8 Partial 10 SubormitHint I give up Hits: 1 % deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback - 50% Part (b) Find the distance, d, in meters. Al content © 2018 Expert TA, LLCExplanation / Answer
Given,
Initial height h = 6.4 m
Length of plane, L = 6.4/sin29 = 13.20 m
Net force, F = ma = mg sin(theta) - u g cos(theta)
Acceleration, a = g sin(theta) - ug cos(theta)
a = (9.8 x sin29) - (0.1 x 9.8 x cos29) = 3.89 m/s^2
Velocity, v = sqrt (2al) = sqrt(2 x 3.89 x 13.20) = 10.13 m/s
B)
Acceleration, a = - u g
Distance travelled, d = u^2/(2 x ug)
d = 10.13^2/(2 x 0.2 x 9.8) = 26.18 m
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