PartA A skier starts down a frictionless 31° slope. After a vertical drop of 22

Question

PartA A skier starts down a frictionless 31° slope. After a vertical drop of 22 m, the slope temporarily levels out and then drops at 15° an additional 38 m vertically before leveling out again. What is the skier's speed on the upper level stretch? Express your answer using two significant figures. VA = 34 m/s Submit Previous Answers Request Answer XIncorrect; Try Again; 3 attempts remaining Part B What is the skier's speed on the lower level stretch? Express your answer using two significant figures. 2 21 m/s Submit Prev X Incorrect. T Again, 3 attempts remaining

Explanation / Answer

During the drop, the skier lost an amount of potential energy equal to m*g*h where h is the vertical height;

Since there is no friction or other dissipating forces, all this PE is converted to KE

Thus we can find speed at the bottom of the first drop by equating initial PE and final KE

mgh = 1/2 mv^2 => v=sqrt[2hg]=sqrt[2x22mx9.8m/s/s]

v = 20.765m/s

when the skier starts down the next slope, there is no friction, hence, KE at the bottom of the slope is equal to the total energy at the top of the slope

1/2mv(bottom)^2= 1/2 mv(top of second slope)^2 + mgh(second drop)

1/2v(bottom)^2 = ½ (20.765m/s)^2 + 9.8m/s/s x 38m

v(bottom)= 34.257m/s

we could also determine the speed at the bottom by considering the trip to be one long vertical drop of 68m; the speed at the bottom of a 60m drop is

v=sqrt[2x9.8m/s/s x 60m]= 34.29m/s

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