a body of mass 2kg is lifted up to a height of 10m & than allowed to fall freely . find the potential & kinetic energy that it has just before touching the ground.
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just before reaching the ground
h=0 m
so potential energy = 0
and initial velocity = 0 so
2gh=v^2
2*10*10=v^2
200=v^2
v=√200m/s
now k.e = 1/2*2*√200*√200
=200 J
h=0 m
so potential energy = 0
and initial velocity = 0 so
2gh=v^2
2*10*10=v^2
200=v^2
v=√200m/s
now k.e = 1/2*2*√200*√200
=200 J
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mass of body, m = 2kg
it is lifted to height = 10m
initial KE = 0
initial PE = mgh = 2×10×10 = 200J
when it is dropped, just before touching ground,
height = 0
final PE = mgh = 2×10×0 = 0J
According to conservation of energy,
Total initial energy = Total final energy
⇒ 0 + 200 = 0 + final KE
⇒ final KE = 200J
So, just before touching ground, KE is 200J and PE is 0J.
it is lifted to height = 10m
initial KE = 0
initial PE = mgh = 2×10×10 = 200J
when it is dropped, just before touching ground,
height = 0
final PE = mgh = 2×10×0 = 0J
According to conservation of energy,
Total initial energy = Total final energy
⇒ 0 + 200 = 0 + final KE
⇒ final KE = 200J
So, just before touching ground, KE is 200J and PE is 0J.
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