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**

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

https://brainly.ph/question/233160

https://brainly.ph/question/78768

- Definition and example of kinetic energy

https://brainly.ph/question/126300

https://brainly.ph/question/77384

- Total mechanical energy of a yoyo

https://brainly.ph/question/1374849

#LetsStudy

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