Question

use the worked example above to help you solve this problem. a 57.0 kg skier is at the top of a slope, as shown in the figure. at the initial point \textcircled{a}, she is 11.0 m vertically above point \textcircled{b}. (a) setting the zero level for gravitational potential energy at \textcircled{b}, find the gravitational potential energy of this system when the skier is at \textcircled{a} and then at \textcircled{b}. finally, find the change in potential energy of the skier - earth system as the skier goes from point \textcircled{a} to point \textcircled{b}. (pe_i=) j (pe_f=) j (delta pe=) j (b) repeat this problem with the zero - level at point \textcircled{a}. (pe_i=) j (pe_f=) j (delta pe=) j (c) repeat again, with the zero level 2.00 m higher than point \textcircled{b}. (pe_i=) j (pe_f=) j (delta pe=) j exercise hints: getting started | im stuck! use the values from practice it to help you work this exercise. if the zero level for the gravitational potential energy is selected to be midway down the slope, 5.50 m above point \textcircled{b}, find the initial potential energy, the final potential energy, and the change in potential energy as the skier goes from point \textcircled{a} to \textcircled{b} in the figure. initial kj final kj

Explanation:

Step1: Recall gravitational - potential - energy formula

The formula for gravitational potential energy is $PE = mgh$, where $m$ is the mass, $g = 9.8\ m/s^{2}$ is the acceleration due to gravity, and $h$ is the height relative to the zero - potential level.

Step2: Calculate for part (a) with zero - level at B

Initial potential energy at A

$PE_i=mgh_i$, with $m = 57.0\ kg$, $g = 9.8\ m/s^{2}$, and $h_i=11.0\ m$. So $PE_i=57.0\times9.8\times11.0 = 6114.6\ J$.

Final potential energy at B

Since $h_f = 0\ m$ (zero - level at B), $PE_f=0\ J$.

Change in potential energy

$\Delta PE=PE_f - PE_i=0 - 6114.6=- 6114.6\ J$.

Step3: Calculate for part (b) with zero - level at A

Initial potential energy at A

Since $h_i = 0\ m$ (zero - level at A), $PE_i = 0\ J$.

Final potential energy at B

$h_f=-11.0\ m$ (negative because B is below A), so $PE_f=57.0\times9.8\times(-11.0)=-6114.6\ J$.

Change in potential energy

$\Delta PE=PE_f - PE_i=-6114.6-0=-6114.6\ J$.

Step4: Calculate for part (c) with zero - level 2.00 m above B

Initial potential energy at A

$h_i = 11.0 - 2.00=9.00\ m$, so $PE_i=57.0\times9.8\times9.00 = 5012.76\ J$.

Final potential energy at B

$h_f=-2.00\ m$, so $PE_f=57.0\times9.8\times(-2.00)=-1117.2\ J$.

Change in potential energy

$\Delta PE=PE_f - PE_i=-1117.2 - 5012.76=-6129.96\ J$.

Step5: Calculate for the exercise with zero - level 5.50 m above B

Initial potential energy at A

$h_i=11.0 - 5.50 = 5.50\ m$, so $PE_i=57.0\times9.8\times5.50=3057.3\ J$.

Final potential energy at B

$h_f=-5.50\ m$, so $PE_f=57.0\times9.8\times(-5.50)=-3057.3\ J$.

Change in potential energy

$\Delta PE=PE_f - PE_i=-3057.3-3057.3=-6114.6\ J$.

Answer:

(a) $PE_i = 6114.6\ J$, $PE_f = 0\ J$, $\Delta PE=-6114.6\ J$
(b) $PE_i = 0\ J$, $PE_f=-6114.6\ J$, $\Delta PE=-6114.6\ J$
(c) $PE_i = 5012.76\ J$, $PE_f=-1117.2\ J$, $\Delta PE=-6129.96\ J$
Exercise: $PE_i = 3057.3\ J$, $PE_f=-3057.3\ J$, $\Delta PE=-6114.6\ J$