Olive Udadi Is At The Park With Her Father. The 26-kg Olive Is On A Swing Following…

Olive Udadi is at the park with her father. The 26-kg Olive is on a swing following the path as shown. Olive has a speed of 0 m/s at position A and is a height of 3.0-m above the ground. At position B, Olive is 1.2 m above the ground. At position C (2.2 m above the ground), Olive projects from the seat and travels as a projectile along the path shown. At point F, Olive is a mere picometer above the ground. Assume negligible air resistance throughout the motion. Use this information to fill in the table. (TME=total mechanical energy)

Answer:

A. PE = 764.4 J , KE = 0 J , TME = 764.4 J

B. PE = 305.76 J , KE = 44,946.72 J, TME = 45,252.48 J

C. PE = 560.56 J , KE = 7191.48 J, TME = 7752.04 J

F. PE = 0J , KE = 44,946.72 J, TME = 44,946.72 J

Explanation:

In this problem, we are to fill a table about the potential and kinetic energy of Olive as well as her total mechanical energy. In computing the total mechanical energy (TME), we will just add PE and KE.

Formula to use

a. PE = mgh

where

PE     is the potential energy, the unit is in Joules (J)

m      is the mass, the unit is in kg

g       is the Earth’s pull of gravity, 9.8 m/s²

h        is the height, the unit is in meter (m)

b. KE = ¹/₂mv²

where

KE     is the kinetic energy

v        is the velocity, the unit is in m/s

c. TME = PE + KE

where

TME     is the total mechanical energy

Solving the information

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Let us solve the following requirements to fill the table using the three given formula.

At point A:

PE = mgh = (26 kg)(9.8 m/s²)(3 m) = 764.4 J

KE = ¹/₂mv² = ¹/₂(26 kg)(0)² = 0 J

TME = PE + KE = 764.4 J + 0 = 764.4 J

At point B:

PE = mgh = (26 kg)(9.8 m/s²)(1.2 m) = 305.76 J

To solve for the velocity at point B, take note that PE at point A is equivalent to KE at point B. Therefore,

PE = KE

mgh = ¹/₂mv²

Simplify

2gh = v²

v = [tex]\sqrt{2gh}[/tex]

substitute the given data

v = [tex]\sqrt{2(9.8)(3)}[/tex]

v = 58.8 m/s

KE = ¹/₂mv² = ¹/₂(26 kg)(58.8 m/s)² = 44,946.72 J

TME = PE + KE = 305.76 J + 44,946.72 J = 45,252.48 J

At point C:

PE = mgh = (26 kg)(9.8 m/s²)(2.2 m) = 560.56 J

To solve for the velocity at point C, take note that PE at point B is equivalent to KE at point C. Therefore,

PEb = KEc

mgh = ¹/₂mv²

Simplify

2gh = v²

v = [tex]\sqrt{2gh}[/tex]

substitute the given data

v = [tex]\sqrt{2(9.8)(1.2)}[/tex]

v = 23.52 m/s

KE = ¹/₂mv² = ¹/₂(26 kg)(23.52 m/s)² = 7191.48 J

TME = PE + KE = 560.56 J + 7191.48 J = 7752.04 J

At point F:

PE = mgh = (26 kg)(9.8 m/s²)(0) = 0 J

To solve for the velocity at point F, take note that PE at point D is equivalent to KE at point F. Therefore,

PEd = KEf

mgh = ¹/₂mv²

Simplify

2gh = v²

v = [tex]\sqrt{2gh}[/tex]

substitute the given data

v = [tex]\sqrt{2(9.8)(3)}[/tex]

v = 58.8 m/s

Let us assume that the height at point D is the same as the height at point A.

KE = ¹/₂mv² = ¹/₂(26 kg)(58.8 m/s)² = 44,946.72 J

TME = PE + KE = 0 J + 44,946.72 J = 44,946.72 J

To learn more, just click the following links below:

  • Definition example of potential energy

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        https://brainly.ph/question/78768

  • Definition and example of kinetic energy
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