Derive an expression for the gravitational potential energy of an object of mass 'm' at a height 'h' above the earth's surface.
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force=ma(mg)
displacement=h
work=energy
=force*displacement
=mg*h
=mgh
i hope it helps u!! mark as brainliest pls!!
displacement=h
work=energy
=force*displacement
=mg*h
=mgh
i hope it helps u!! mark as brainliest pls!!
Answered by
0
Suppose I let an object fall from rest for a distance hh. Given that the gravitational acceleration is gg the velocity of the object will be given by the SUVAT equation:
v2=u2+2gh
v2=u2+2gh
In this case the initial velocity u=0u=0 so we just get v2=2ghv2=2gh. The kinetic energy of the object is given by:
T=12mv2=12m(2gh)=mgh
T=12mv2=12m(2gh)=mgh
If energy is conserved the increase in kinetic energy must be equal to the decrease in potential energy, so we get:
ΔU=−ΔT=−mgh
ΔU=−ΔT=−mgh
This tells us that if we lower the object by a distance hh the potential energy decreases by mghmgh, and conversely that if we raise it by a distance hh the potential energy increases by mghmgh.
v2=u2+2gh
v2=u2+2gh
In this case the initial velocity u=0u=0 so we just get v2=2ghv2=2gh. The kinetic energy of the object is given by:
T=12mv2=12m(2gh)=mgh
T=12mv2=12m(2gh)=mgh
If energy is conserved the increase in kinetic energy must be equal to the decrease in potential energy, so we get:
ΔU=−ΔT=−mgh
ΔU=−ΔT=−mgh
This tells us that if we lower the object by a distance hh the potential energy decreases by mghmgh, and conversely that if we raise it by a distance hh the potential energy increases by mghmgh.
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