Skier down a slope and potential energy

[rkslperez04]

Here is the question:

A women skis down a slope 100 m high. Her speed at the foot of the slope is 20 m/s. What percentage of her intial potential engery is lost?


okkk.. here is what I get:

My instructor mentioned this was a Ef/Ei problem. (effientcy final divided by effientcy intial)

my book uses the formula: 1 - (KE2/PE2)

Im confused where the one came from?

then...

I understand we have PE at the top of the hill only being we are to assume she is at rest... so no KE. So does that mean we no PE at the bottom only KE...


Im confused.. can someone explain this to me in laymens terms so I can rework the problem. Is there an easier way to solve this.??

 

[PhanthomJay]

Here is the question:

A women skis down a slope 100 m high. Her speed at the foot of the slope is 20 m/s. What percentage of her intial potential engery is lost?


okkk.. here is what I get:

My instructor mentioned this was a Ef/Ei problem. (effientcy final divided by effientcy intial)

my book uses the formula: 1 - (KE2/PE2)

Im confused where the one came from?

then...

I understand we have PE at the top of the hill only being we are to assume she is at rest... so no KE. So does that mean we no PE at the bottom only KE...


Im confused.. can someone explain this to me in laymens terms so I can rework the problem. Is there an easier way to solve this.??

Yes, you are on the right track. She has only PE at the top. At the bottom, no PE, just KE. Calculate her PE at the top. Calculate her KE at the bottom. How much energy was lost (due to friction, snow resistance, air resistance, etc.)? What percentage is that of the initial energy? The result is the same as (1 - E_f/E_i), where E_f and E_i are the final and initial energy, respectively. I don't understand either where the book formula came from.

 

[andrevdh]

The amount of energy that was lost is

PE_t - KE_b

where t refers to top and b to bottom. The fraction of energy lost will be

\frac{PE_t - KE_b}{PE_t}

which comes to

1 - \frac{KE_b}{PE_t}

the percentage will just be this fraction times one hundred.