A 45 kg skateboarder starts from rest at the top of the track in the figure. he

Question

A 45 kg skateboarder starts from rest at the top of the track in the figure. he begins a decent down the track, always maintaining contact with the surface. The mass of the skateboard is negligible, as is friction, except where noted.
a) what is the skateboarders speed when he reaches the bottom of the initial dip, 12.0 m below the starting point?

b) What is his speed when he reaches the highest point (8.0 m, labeled) on the other side of the dip?

c) as he begins to descend again, down a straight, 18.0 m slope, he slows his skateboard down by using friction on the tail of the board. He is able to produce a friction force with a magnitude of 120.0 N. What is his speed when he reaches the bottom? (end of 18.0 m length of track)?

ra' in the figare. He begins a descent down the he mass of the skaicbosrd is negligible, as is friction, n hi: reaches the vocvom of the mitisal dip, 12.l m below

Explanation / Answer

(a) Let v be the speed at the bottom of initial dip.

Total energy is conserved.

KE + PE = constant

=> mgh + 0 = 0 + mv2/2

=> v = (2gh)1/2 = (2 * 9.8 * 12)1/2 = 15.34 m/s

b) Let v1 be the required speed. Then, energy conservation gives.

0 + m*15.342/2 = mg*8 + mv12/2

=> v1 = [15.342 - 2*9.8*8]1/2 = 8.86 m/s

c) Let be the angle of the slope.

Then, sin = 8/18 = 0.44

Accceleration, a = [mg(sin ) - f]/m = [45*9.8*0.44 - 120]/45 = 1.65 m/s2

Final speed at bottom, v2 = [8.862 + 2*1.65*18]1/2 = 11.74 m/s

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