A skier approaches a snowy hill moving at 13.9 m/s. The coefficients of static a

Question

A skier approaches a snowy hill moving at 13.9 m/s. The coefficients of static and kinetic friction between the snow and the skies are 0.450 and 0.310, respectively, and the hill slopes upward at 44.0 degrees above the horizontal. Find the acceleration of the skier while going up the hill. Take uphill as the positive direction. How far will the skier travel along the plane before he stops? What height above the ground will he reach? Once the skier reaches the location where he instantaneously stops, will he stay there or begin to slide downhill.

Explanation / Answer

A)    Here, net acceleration of skier = g * sin(theta) + uk * g * cos(theta)

                                                      =    9.8 * sin(44) + 0.310 * 9.8 * cos(44)

                                                      =   - 8.993 m/sec2

B)     Here,    1/2 * m * 13.92 = m * (9.8 * sin(44) + 0.310 * 9.8 * cos(44)) * x

=>      x   = 10.742 m

=>    Distance skier travel along plane = 10.742 m

=>    Height above ground = 10.742 * sin44

                                            = 7.462 m

C)   At point of stopping

=>   Force of static friction = 0.450 * m * 9.8 * cos44   = 3.172m

=>   Gravitational force = m * 9.8 * sin44 = 6.8m

=> skier begin to slide downhill .

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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