Answer:
there is a net loss of potential energy is Ut = -0.11 10⁵ J
Explanation:
The gravitational potential energy is
ΔU = mg (y2- y1)
Where y1 is the initial position where the energy is defined as zero, in general in problems on the Earth's surface the reference system is taken at the lowest point, whereby y1 = 0 and consequently U1 = 0, this It is a matter of comfort since what matters are the changes in energy and not its value at one point explained.
Let's apply these concepts to our problem.
Jump from the ground to 179 m, let's calculate the energy change
ΔU1 = mg (y2-y1)
ΔU1 = 67.5 9.8 (179 -0)
ΔU1 = 1.18 10⁵ J gains energy
Dives up to Y3 = -15.9m below ground, therefore, this height is negative
The total height is y = 179 - (-15.9) = 194.9 m
ΔU2 = mg (y3-y2)
These are the points where the dive begins and where it arrives
ΔDU2 = 67.5 9.8 (- 194.9)
ΔU2 = -1.29 105 J loses energy
The total change in potential energy is
Ut = ΔU1 + ΔU2
Ut = 1.18 10⁵ - 1.29 10⁵
Ut = -0.11 10⁵ J
Therefore, there is a net loss of potential energy
You can also calculate ΔU = mg (Δy)
Where ΔY is the change in body position, that is -15.9 m
ΔU = 67.5 9.8 (-15.9)
ΔU = -1.1 10⁴ J = -0.11 10⁵ J
You can see that it is the same value.
