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?
Answer
Mass of object m=60kg
Raised at height h=12m
So potential energy
U=mgh
"=60\\times 9.81\\times12\\\\=7063.2J"
Potential energy at height h'=8m(two-thirds)
"U'=mgh'=60\\times9.81\\times8"
"\\\\=4708.8J"
Let kinetic energy is E at this height
Now using energy conservation
"Ui+Ei=Uf+Ef\\\\7063.2+0=4708.8+E\\\\E=2354.4J"
So velocity of object
"v=\\sqrt{\\frac{2E}{m}}\\\\=\\sqrt{\\frac{2\\times2354.4}{60}}\\\\=8.86\\space m\/s"
Comments
Leave a comment