A 75 kg skier has a speed of 8.4 m/s at point A. She then encounters a 1.75 m de

Question

A 75 kg skier has a speed of 8.4 m/s at point A. She then encounters a 1.75 m deep dip in the snow's surface that has a circular cross section with a radius of curvature of 12 m.
Note: You may ignore friction in this problem.

a.) How large is the normal force exerted by the snow on the skier at point B?
N (previous INCORRECT answers have been 458, 1176, 13902, 735.75). I found 1390 but have not input it as I need to confirm it is correct as it is my last chance.

b.) How much work is done by the normal force on the skier from point A to point B?
0 J

I have solved this 5 times and need to find the correct answer

I believe this may be right, but need to confirm

this is how i solved....

use the below kinematic equation, we get

v^2 - u^2 = 2 *a d

v^2 - 8.4^2 = 2 * 9.8 * 1.75

v = 10.24 m/s

at the bottom of the dip, normal force at the bottom of the dip is

N = mg + mv^2/R = 75 *(10.24^2/12 + 9.8) = 1390. 375 N = 1390 N

b.

the normal force is always perpendicular to the direction of motion

Work done by normal force = Fdcos90o = 0 J

Explanation / Answer

N=1391.34 Newton

Work done = 0

U r right.

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