shyam drops a ball of 100gm.from a building of height 10m .what will be its kinetic energy at the height of 4m? what will happen to its total mechanical energy? give reasons to justify your answer.
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Given, mass, m = 100 g= 0.1 kg, height, h = 10 m
Potential energy, PE = mgh = 0.1 X10 X10 = 10J
When body reaches at height 6 m , in velocity is given by
${{v}^{2}}$ = ${{u}^{2}}$ + 2gh
${{v}^{2}}$ = 0 + 2 x 10 x 6

Now, its kinetic energy at this point is given by,
= 1/2 m${{v}^{2}}$ = 1/2 x 0.1 x 4 x 30 = 6 J
As, mechanical energy = KE + PE [at highest point]
Mechanical energy = 0+10 J = 10J
Mechanical energy at 4 m height
= KE+ PE= 6 + 0.l x 10x 4= 10J
Hence, it is seen that total mechanical energy always remains constant during the motion.
Potential energy, PE = mgh = 0.1 X10 X10 = 10J
When body reaches at height 6 m , in velocity is given by
${{v}^{2}}$ = ${{u}^{2}}$ + 2gh
${{v}^{2}}$ = 0 + 2 x 10 x 6

Now, its kinetic energy at this point is given by,
= 1/2 m${{v}^{2}}$ = 1/2 x 0.1 x 4 x 30 = 6 J
As, mechanical energy = KE + PE [at highest point]
Mechanical energy = 0+10 J = 10J
Mechanical energy at 4 m height
= KE+ PE= 6 + 0.l x 10x 4= 10J
Hence, it is seen that total mechanical energy always remains constant during the motion.
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