A 64.0-kg skier starts from rest at the top of a ski slope of height 62.0 m . Pa

Question

A 64.0-kg skier starts from rest at the top of a ski slope of height 62.0 m .

Part A

If frictional forces do 1.05×104 J of work on her as she descends, how fast is she going at the bottom of the slope?

Take free fall acceleration to be g = 9.80 m/s2

Part B

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.23. If the patch is of width 69.0 m and the average force of air resistance on the skier is 170 N , how fast is she going after crossing the patch?

Part C

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.4 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

given,

mass = 64 kg

height = 62 m

frictional force = -1.05 * 10^4 J

by conservation of energy

initial energy = final energy

mgh = 0.5 * mv^2 + frictional force

64 * 9.8 * 62 = 0.5 * 64 * v^2 + 1.05 * 10^4

speed at the botton of the slope v = 29.7838 m/s

0.5 * 64 * 29.7838^2 = 0.5 * 64 * v^2 + 0.23 * 64 * 9.8 * 69 + 170 * 69

speed after crossing the patch v = 14.4727 m/s

0.5 * 64 * 14.4727^2 = F * 2.4

force exerted by snowdrift F = 2792.79 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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