Body mass 5kg thrown up vertically with KE 1000J. If acceleration due to gravity is 10ms. Find height at which KE become half original value _
Answers
So, this is an issue of the conservation of energy, which states…
ΔEsys+ΔEelse=0
Where ΔE means change in energy, and sys and else refer to the system and the rest of the universe.
If we ignore air resistance, the ΔEelse term becomes zero. So we get…
ΔEsys=0
We know that ΔE=ΔU+ΔKE+ΔPE
Where U, KE, and PE are Internal Energy, Kinetic Energy, and Potential Energy respectively.
We will assume that the Internal Energy is not changing, which means…
ΔE=ΔKE+ΔPE
Since ΔE also equals 0
ΔKE+ΔPE=0→ΔKE=−ΔPE
Now we’ve made all of our assumptions/simplifications, but there’s one last thing…
In a uniform gravitational field (which earth gravity mostly is for fairly low heights), you find that ΔPE=mgΔh where m is mass, g is gravitational field (acceleration due to gravity), and h is change in height.
Now, we want to find Δh where ΔKE=−12KE0 where KE0 is some arbitrary initial value. So our final equation is this
−12KE0=−mgΔh→Δh=12mgKE0
Now we just plug in all of the numbers for the specific situation the question asked for (which we know)
Δh=12mgKE0=12(5kg)(9.8ms2)(490J)=5m
Where all units are SI (so your final answer is in meters)
Hopefully this helps
