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# A skier is skiing down a frictionless 30 degree slope. She starts at rest. Her mass is 50 kg.?

a. What will her acceleration be along the slope?

b. What will be the normal force that the slope exerts on her?

c. If the coefficient of static friction is 0.20, will she slide down the hill?

### 2 Answers

- oubaasLv 71 month agoFavorite Answer
a

acceleration a = g*sin 30° = 4.903 m/sec^2

b

N = m*g*cos 30° = 50*9.806*0.866 = 425 N

c

motive force MF = m*g*sin 30°

friction force FF = m*g*cos 30°*μ

accelerating force AF = MF-FF = m*g*(sin 30°-cos 30°*μ )

acceleration a' = AF/m = g*(sin 30°-cos 30°*μ ) = 9.806*(0.5-0.173) = 3.2 m/sec^2

..she will slide down the hill with an acceleration a' = 3.2 m/sec^2

- billrussell42Lv 71 month ago
if she stopped, her velocity and acceleration are zero

weight = 9.8x50 = 49 N

component perpendicular to slope = 49cos30 = 49(√3/2) N = 42.4 N

component parallel to slope = 49sin30 = 24.5 N

force of friction = 0.2•49(√3/2) = 8.49 N

Yes, force of friction is less than 24.5 N