QUESTION If the angle of the slope is increased, the change of gravitational potential energy between the two heights: (Select all that apply.) O remains the same. O depends only on the difference between the two heights. sometimes decreases and sometimes increases. depends on the path followed. increases. decreases.

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Chapter1: Units, Trigonometry. And Vectors
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Please answer the final question at the bottom.
PROBLEM A 60.0 kg skier is at the top of a slope, as
shown in the figure. At the initial point O, she is 10.0 m
vertically above point ®. (a) Setting the zero level for
gravitational potential energy at ®, find the gravitational
potential energy of this system when the skier is at O and
then at ®. Finally, find the change in potential energy of the
skier-Earth system as the skier goes from point O to point
®. (b) Repeat this problem with the zero level at point
O. Repeat again, with the zero level 2.00 m higher than
10.0 m
point ®
STRATEGY Follow the definition and be careful with signs.
O is the initial point, with gravitational potential energy PE,,
and ® is the final point, with gravitational potential energy PEf. The location chosen for y = 0 is also the
zero point for the potential energy, because PE = mgy.
SOLUTION
(A) Let y
= 0 at ®. Calculate the potential energy at O and at ®, and calculate the change in potential
energy.
Find PE,, the potential energy at
O.
PE, = mgy; = (60.0 kg)(9.80 m/s²)(10.0 m) =
5.88 x 103 )
PE, = 0 at ® by choice. Find the
PE, - PE, = 0 - 5.88 x 103 ) = -5.88 x 10³ J
difference in potential energy between
O and O
(B) Repeat the problem if y = 0 at O, the new reference point, so that PE = 0 at O.
Find PE, noting that point ® is now at
PE, = mgy, = (60.0 kg)(9.80 m/s²)(-10.0 m) =
y = -10.0 m.
-5.88 x 103 J
PE, - PE, = -5.88 x 103 J -0 = -5.88 x 10³ J
(C) Repeat the problem, if y = 0 two meters above ®.
Find PE, the potential energy at O.
PE, = mgy, = (60.0 kg)(9.80 m/s2)(8.00 m) =
4.70 x 103 J
Find PE, the potential energy at ®.
PE, = mgy, = (60.0 kg)(9.80 m/s²)(-2.00 m) =
-1.18 x 103 )
Compute the change in potential
PE, - PE, = -1.18 × 103 J - 4.70 × 10³ ) =
energy.
-5.88 x 103 )
LEARN MORE
REMARKS These calculations show that the change in the gravitational potential energy when the skier
goes from the top of the slope to the bottom is -5.88 x 10' ), regardless of the zero level selected.
QUESTION If the angle of the slope is increased, the change of gravitational potential energy between
the two heights: (Select all that apply.)
O remains the same.
O depends only on the difference between the two heights.
O sometimes decreases and sometimes increases.
O depends on the path followed.
O increases.
O decreases.
Transcribed Image Text:PROBLEM A 60.0 kg skier is at the top of a slope, as shown in the figure. At the initial point O, she is 10.0 m vertically above point ®. (a) Setting the zero level for gravitational potential energy at ®, find the gravitational potential energy of this system when the skier is at O and then at ®. Finally, find the change in potential energy of the skier-Earth system as the skier goes from point O to point ®. (b) Repeat this problem with the zero level at point O. Repeat again, with the zero level 2.00 m higher than 10.0 m point ® STRATEGY Follow the definition and be careful with signs. O is the initial point, with gravitational potential energy PE,, and ® is the final point, with gravitational potential energy PEf. The location chosen for y = 0 is also the zero point for the potential energy, because PE = mgy. SOLUTION (A) Let y = 0 at ®. Calculate the potential energy at O and at ®, and calculate the change in potential energy. Find PE,, the potential energy at O. PE, = mgy; = (60.0 kg)(9.80 m/s²)(10.0 m) = 5.88 x 103 ) PE, = 0 at ® by choice. Find the PE, - PE, = 0 - 5.88 x 103 ) = -5.88 x 10³ J difference in potential energy between O and O (B) Repeat the problem if y = 0 at O, the new reference point, so that PE = 0 at O. Find PE, noting that point ® is now at PE, = mgy, = (60.0 kg)(9.80 m/s²)(-10.0 m) = y = -10.0 m. -5.88 x 103 J PE, - PE, = -5.88 x 103 J -0 = -5.88 x 10³ J (C) Repeat the problem, if y = 0 two meters above ®. Find PE, the potential energy at O. PE, = mgy, = (60.0 kg)(9.80 m/s2)(8.00 m) = 4.70 x 103 J Find PE, the potential energy at ®. PE, = mgy, = (60.0 kg)(9.80 m/s²)(-2.00 m) = -1.18 x 103 ) Compute the change in potential PE, - PE, = -1.18 × 103 J - 4.70 × 10³ ) = energy. -5.88 x 103 ) LEARN MORE REMARKS These calculations show that the change in the gravitational potential energy when the skier goes from the top of the slope to the bottom is -5.88 x 10' ), regardless of the zero level selected. QUESTION If the angle of the slope is increased, the change of gravitational potential energy between the two heights: (Select all that apply.) O remains the same. O depends only on the difference between the two heights. O sometimes decreases and sometimes increases. O depends on the path followed. O increases. O decreases.
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