5. As she skis down a slope that makes an angle of 15 degree with the horizontal

Question

5. As she skis down a slope that makes an angle of 15 degree with the horizontal, a 50-kg skier hits a sunny region where the snow has melted and has become sticky. As a result, she slows down at a rate of 3 m/s^2 in this region. i. Find the coefficient of friction between the skis and the slope in this region. ii. The skier's speed was 22 m/s at the instant she entered the sunny region. Find her speed when she leaves the sunny region 4 seconds later. iii. Find the distance that the skier travels while she is in the sunny region.

Explanation / Answer

i) force equation

mgsin(theta) - friction = ma

m*9.8*sin15 - u*N = ma

m*9.8*sin15 - u*mgcos(theta) = m*(-3)

9.8*sin15 - u*9.8*cos15 = -3

2.536 + 3 = u*9.466

u = 0.5848

ii) v = u + at

v = 22 - 3*4

v = 10 m/s

iii) s = ut+0.5*at^2

s = 22*4 + 0.5*(-3)*4^2

s = 64 m

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.