Answered

IDNLearn.com connects you with a community of experts ready to answer your questions. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.

A box of mass 3.6 kg is lifted 5.4 m above the
ground. What is the change in energy in lifting
the box to this height?

Sagot :

MDKPfromIDNidn

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.

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.