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A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the coconut's potential and kinetic energy at each point shown in the picture. Its speed is z…

A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the coconut's potential and kinetic energy at each point shown in the picture. Its speed is z…
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A 1-kilogram coconut falls out of a tree from a height of 12 meters.

Determine the coconut's potential and kinetic energy at each point
shown in the picture. Its speed is zero at point A.
A
3m
B
3m
с
3m
D
3m
E

asked
User Steve Faro
by
8.0k points

1 Answer

1 vote

\begin{gathered} mass=\text{ m= 1kg} \\ h_A=12\text{ m} \\ At\text{ point A} \\ E_A=E_(PA)+E_{KA,\text{ }}\text{ }E_(PA)=mgh_A\text{ , }E_{KA,\text{ }}=(mv^2_A)/(2) \\ E_A=mgh_A+(mv^2_A)/(2) \\ v_A=\text{ 0 m/s, Then} \\ E_A=E_(PA)=mgh_A \\ E_(PA)=\mleft(1\operatorname{kg}\mright)(9.81_{}m/s^2)(12m) \\ E_(PA)=117.72J \\ \text{The potential energy at point A is 117.72 J and kinetic energy is 0J} \\ \\ At\text{ point B} \\ E_A=E_B \\ E_(PA)=E_(PB)+E_{KB,\text{ }} \\ mgh_A=mgh_B+E_{KB,\text{ }} \\ E_(PB)=mgh_B \\ h_B=9m \\ E_(PB)=(1\operatorname{kg})(9.81_{}m/s^2)(9m) \\ E_(PB)=88.29J,\text{ then} \\ 117.72J=88.29J+E_{KB,\text{ }} \\ E_{KB,\text{ }}=117.72J-88.29J \\ E_{KB,\text{ }}=29.43J \\ \text{The potential energy at point B is }88.29J\text{ and kinetic energy is }29.43J \\ \\ At\text{ point C} \\ E_B=E_C \\ E_(PB)+E_(KB)=E_(PC)+E_(KC) \\ 117.72J=mgh_C+E_{KC\text{ }} \\ E_(PC)=mgh_C \\ E_(PC)=(1\operatorname{kg})(9.81_{}m/s^2)(6m) \\ E_(PC)=58.86J,\text{ then} \\ 117.72J=58.86J+E_{KC\text{ }} \\ E_{KC\text{ }}=117.72J-58.86J \\ E_{KC\text{ }}=58.86J \\ \text{The potential energy at point C is }58.86J\text{ and kinetic energy is }58.86J \\ \\ At\text{ point D} \\ E_C=E_D \\ E_(PC)+E_(KC)=E_(PD)+E_(KD) \\ 117.72J=mgh_D+E_{KD\text{ }} \\ E_(PD)=mgh_D \\ E_(PD)=(1\operatorname{kg})(9.81_{}m/s^2)(3m) \\ E_(PD)=29.43J \\ \text{Then} \\ 117.72J=29.43J+E_{KD\text{ }} \\ E_{KD\text{ }}=117.72J-29.43J \\ E_{KD\text{ }}=88.29J \\ \text{The potential energy at point D is }29.43J\text{ and kinetic energy is }88.29J \\ \\ At\text{ point E} \\ E_D=E_E \\ E_(PD)+E_(KD)=E_(PE)+E_(KE) \\ 117.72J=mgh_E+E_{KE\text{ }} \\ E_(PE)=mgh_E \\ E_(PE)=(1\operatorname{kg})(9.81_{}m/s^2)(0m) \\ E_(PE)=0J \\ \text{Then} \\ 117.72J=E_{KE\text{ }} \\ \text{The potential energy at point E is }0J\text{ and kinetic energy is 117.72}J \end{gathered}

A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the-example-1
A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the-example-2
A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the-example-3
A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the-example-4
A 1-kilogram coconut falls out of a tree from a height of 12 meters. Determine the-example-5
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User BSAFH
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