Hayden C Dropbox, Inc (US) https://www. dropbox.com/home/phys%20209%20t2?preview

Question

Hayden C Dropbox, Inc (US) https://www. dropbox.com/home/phys%20209%20t2?preview 20170408 144939 jpg EE: Apps Q ART 100 Exam 2 Flash w art WebAssign LOG IN PAWS E oo LL Boas and Anacondas Vertical Motion Phy YouTube N Netflix Moodle e Books Home Dropbox My Notes Ask Your contabt wn the track, always maintaini l skato perk. He starus from rpst at je top ffhe track gligible, as is friction except whv're notod with the surface. The mass of the skateboa 8.0 m eaches the bottom of the ir vilial dip, 12.0 m below the starting point oucends ouwr side of the dip to the top of a hill, 8.0 m above the ground what his speed when he rear this point? descend again, down a straight, 18.0 m-long slope, he s his skateboard down by using friction on the tail of the board. He is able to produce a friction force with a What is the change in thermal energy of the board rack system as he descends the 18.0-m length of track? Wh when he reaches the bottoon (the end of the 18.0 m length of track)? 2:49 PM ENG 6:21 PM 4/8/2017

Explanation / Answer

from law of conservation of energy

total energy at 12 m = tota energy at top

k1 + U1 = K3 + U3

0 + m*g*h1 = (1/2)*m*v3^2 + m*g*h3

v3 = sqrt(2*g*(h1-h3)

v3 = sqrt(2*9.8*(12-8)) = 8.85 m/s

===========================

change in thermal energy = work done by friction

change in thermal energy = f*L = 111.2*18 = 20001.6 J

========================

total mechanical   energy at top - Wfriction = total mechanical energy at bottom

K1 + U1 + Wf = K4 + U4

0 + m*g*h1 - f*L = (1/2)*m*v4^2 + 0

(44*9.8*12) - (111.2*18) = (1/2)*44*v4^2

v4 = 12 m/s <<<-------answer

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