LEARN MORE Remarks Substituting the symbolic expression vB V2gh into the equatio

Question

LEARN MORE Remarks Substituting the symbolic expression vB V2gh into the equation for the distance d shows that d is linearly proportional to h: Doubling the height doubles the distance traveled. Question Give two reasons why skiers typically assume a crouching position when going down a slope. (Select all that apply.) The acceleration of gravity g is increased by crouching Crouching decreases the mass of the skier. Crouching decreases the skier's inertia Air resistance is lower in the crouched position The ski poles push more effectively when they are more nearly horizontal PRACTICE IT Use the worked example above to help you solve this problem. A skier starts from rest at the top of a frictionless incline of height 20.0 m, as shown in the figure. As the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.203 (a) Find the skier's speed at the bottom. m/s 19.8 (b) How far does the skier travel on the horizontal surface before coming to rest? 390 EXERCISE HINTS GETTING STARTED I I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.203. (The incline makes an angle 6 20.0° with the horizontal.) Need Help Talk to a Tutor

Explanation / Answer

First Problem:

Only fourth option - air resistance is lower in the crouched position, is the correct position.

Explanation -

In crouched position, surface area of the skier is less so less air friction.

Second Problem -

Velocity of the skier at the bottom, v = sqrt[2gh] = sqrt[2*9.81*20] = 19.8 m/s

retardation due to friction, a = mu*g = 0.203*9.81 = 1.99 m/s^2

So, horizontal travelled by the skier before stopping -

s= v^2 / (2*a) = 19.8^2 / (2*1.99) = 98.5 m

Third problem -

In this condition, the downward acceleration, a = g*sin20 - mu*g*cos20 = 9.81*0.34 - 0.203*9.81*0.94

= 9.91*(0.34 - 0.19) = 1.49 m/s^2

Length of the slope, l = 20 / sin20 = 58.48 m

So, the velocity of the skier at the bottom, V = sqrt[2*1.49*58.48] = 13.2 m/s

So, the horizontal distance covered by the skier in this case -

S = V^2 / (2*a) = 13.2^2 / (2*1.99) = 43.8 m

[Please note that the acceleration at the horizontal surface will remain the same as before because the coeff. of friction between the skier and the snow is the same, 0.203]

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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