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1. a 63-kg skier coasts up a snow-covered hill that makes an angle of 25 degrees with the horizontal. the initial speed of the skier is 6.6 m/s. after coasting a distance of 1.9 m up the slope, the speed of the skier is 4.4 m/s. [A] find the work done by the kinetic frictional force that acts on the skis. [B] what is the magnitude of the kinetic frictional force?


2. a water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. as the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.500 s after leaving the slide. ignoring friction and air resistance, find the height H in the drawing.

* sorry about the second question. i cant really draw the picture. but the answer for that question is 6.33 m. i dont know how they got that. :[[

2007-10-28 15:42:42 · 1 answers · asked by Amy C 1 in Science & Mathematics Physics

1 answers

The starting Kinetic is equal to the final kinetic minus work increase in PE and loss due to friction

.5*m*vi^2=.5*m*vo^2-m*g*1.9*sin(25)-f*1.9

f is the frictional force and f*1.9 is the work.

.5*63*6.6^2=.5*63*4.4^2-63*9.81*1.9*sin(25)-f*1.9

solve for f [B] and multiply by 1.9 [A]

2. I assume H is the height of the slide?
If a person travels 5 m is 0.5 s, then vx=10 m/s
Using energy

.5*m*10^2=m*g*H
H=50/9.81
H=5.1 m

Not the H you're looking for.

Is H the height of the slide off the water?

y(t)=.5*9.81*.5^2
=1.22 m

Ah, H is the height of the slide plus the height off the water
5.1+1.22=6.32 m

j

2007-10-29 07:37:49 · answer #1 · answered by odu83 7 · 0 0

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