Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
So, the energy change that occurs is 190.512 J.
Introduction
Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy ([tex] \sf{\Delta PE} [/tex]) can be realized in the equation :
[tex] \sf{W = \Delta PE} [/tex]
[tex] \sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2} [/tex]
[tex] \boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}} [/tex]
With the following condition :
- W = work of subject (J)
- [tex] \sf{\Delta PE} [/tex] = change of potential energy (J)
- m = mass (kg)
- g = acceleration of the gravity (m/s²)
- [tex] \sf{h_2} [/tex] = final height (m)
- [tex] \sf{h_1} [/tex] = initial height (m)
Problem Solving
We know that :
- m = mass = 3.6 kg
- g = acceleration of the gravity = 9.8 m/s²
- [tex] \sf{h_2} [/tex] = final height = 5.4 m
- [tex] \sf{h_1} [/tex] = initial height = 0 m
What was asked :
- W = work of subject = ... J
Step by step :
[tex] \sf{W = \Delta PE} [/tex]
[tex] \sf{W = m \cdot g \cdot (h_2 - h_1)} [/tex]
[tex] \sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)} [/tex]
[tex] \sf{W = 3.6 \cdot 9.8 \cdot 5,4} [/tex]
[tex] \boxed{\sf{W = 190.512 \: J}} [/tex]
So, the energy change that occurs is 190.512 J.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.