A 40kg block is resting at a height of 5m off the ground.If the block is released and fall to the ground, what is the total energy at a height of 2m?
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0
total energy means kinetic energy + potential energy...
so potential energy only depends upon height ..
potential energy= mgh
=40×10(5-2)
=40×10×3
=1200 joule
now for kinetic energy we have to find speed at height 3m
so ..applying v^2= u^2+2gh
initial velocity (u) is zero becoz body is released from rest.
v^2 = 0+ 2×10×3
= 600
now kinetic energy = mv^2/2
=40×600/2
1200 joule
so, Total energy = kinetic energy+potential energy
=1200+1200 =2400 joule..
so potential energy only depends upon height ..
potential energy= mgh
=40×10(5-2)
=40×10×3
=1200 joule
now for kinetic energy we have to find speed at height 3m
so ..applying v^2= u^2+2gh
initial velocity (u) is zero becoz body is released from rest.
v^2 = 0+ 2×10×3
= 600
now kinetic energy = mv^2/2
=40×600/2
1200 joule
so, Total energy = kinetic energy+potential energy
=1200+1200 =2400 joule..
Answered by
3
Total energy = Kinetic + Potential , As there is negligible air resistance mentioned in question, so no energy disspation is considered due to that. Initially, PE = mgH = 40*5*10 J = 2000 J .
Explanation:
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