- #1

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a. How far down the slope does he go before stopping?

I know the answer is 144m, but how do I get there?

b. How much mechanical energy is lost in this process?

1/2(3.06)(20)^2 + (3.06)(9.8)(72)

612+ 2159.136 = 2771.136J lost

- Thread starter ScoutFCM
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- #1

- 17

- 0

a. How far down the slope does he go before stopping?

I know the answer is 144m, but how do I get there?

b. How much mechanical energy is lost in this process?

1/2(3.06)(20)^2 + (3.06)(9.8)(72)

612+ 2159.136 = 2771.136J lost

- #2

HallsofIvy

Science Advisor

Homework Helper

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-gm cos(30) perpendicular to the slope- the push of the slope upward which opposes the force perpendicular to the slope- and the friction force which opposes the down slope force and 0.74 mg cos(30). The net force, down the slope is -mg sin(30)+ 0.74 mg cos(30)= mg(.74 cos(30)-sin(30))= 0.14mg= 1.38m.

The acceleration due to that is F/m= 1.38 m/s

Since the acceleration is a constant and the initial speed was 20 m/s down the slope, the speed at time t is 1.38t- 20 m/s. The distance down the slope at time t is 0.69t

The skier stops when her speed is 0: 1.38t- 20= 0 or t= 20/1.38= 14.5 seconds. In that time she has gone (0.69)(14.5

"b. How much mechanical energy is lost in this process?

1/2(3.06)(20)^2 + (3.06)(9.8)(72)

612+ 2159.136 = 2771.136J lost"

Where did the "3.06" come from? You didn't tell us the mass of the skier!

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