Question #291844

an object of mass 60 kg is raised 12 m and then released. assuming no losses due to air resistance, what is its velocity two-thirds of the way down?


1
Expert's answer
2022-01-30T13:52:00-0500

Answer

Mass of object m=60kg

Raised at height h=12m

So potential energy

U=mgh

=60×9.81×12=7063.2J=60\times 9.81\times12\\=7063.2J

Potential energy at height h'=8m(two-thirds)

U=mgh=60×9.81×8U'=mgh'=60\times9.81\times8

=4708.8J\\=4708.8J

Let kinetic energy is E at this height

Now using energy conservation

Ui+Ei=Uf+Ef7063.2+0=4708.8+EE=2354.4JUi+Ei=Uf+Ef\\7063.2+0=4708.8+E\\E=2354.4J

So velocity of object

v=2Em=2×2354.460=8.86 m/sv=\sqrt{\frac{2E}{m}}\\=\sqrt{\frac{2\times2354.4}{60}}\\=8.86\space m/s



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS