State and prove the law of conservation of energy in the case of a freely falling object.
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⏹Let's consider a case....
✔️Suppose a body is at point A initially. Then it falls freely, during the motion mechanical energy of the body is conserve.
✔️⏹✔️At point A,
➖AC = height of object from ground
=h
➖Initial speed u = 0
➖So, P.E = mgh
K.E=1/2mv2
=1/2m(0)
=0
➖TE = P.E + K.E = mgh + 0
⏹TE= total energy
⏹PE=potential energy
⏹KE= kinetic energy
✔️(TE)A = mgh .................................... (i)
✔️⏹✔️At point B,
PE = mg (BC)
= mg (h – x) = mgh – mgx
Use kinematics equation......
⏹ Refer attachment for formula ⏹
✔️(TE)B = P.E. + K.E = mgh – mgx + mgx
= mgh ...(ii)
✔⏹️✔️At point C
P.E. = 0
KE=1/2mv2
✔️Now use kinematic equation again...
v2=u2+2gh
v 2 = 2gh
So,
➖Substitute the value again...
KE=1/2m(2gh)
=mgh
✔️(T.E.)C = K.E + P.E. = mgh +0 = mgh ...(iii)
✔️(TE)A =(TE)B=(TE)C
Hence, the total energy of the body is conserve during free fall..........
✔️Suppose a body is at point A initially. Then it falls freely, during the motion mechanical energy of the body is conserve.
✔️⏹✔️At point A,
➖AC = height of object from ground
=h
➖Initial speed u = 0
➖So, P.E = mgh
K.E=1/2mv2
=1/2m(0)
=0
➖TE = P.E + K.E = mgh + 0
⏹TE= total energy
⏹PE=potential energy
⏹KE= kinetic energy
✔️(TE)A = mgh .................................... (i)
✔️⏹✔️At point B,
PE = mg (BC)
= mg (h – x) = mgh – mgx
Use kinematics equation......
⏹ Refer attachment for formula ⏹
✔️(TE)B = P.E. + K.E = mgh – mgx + mgx
= mgh ...(ii)
✔⏹️✔️At point C
P.E. = 0
KE=1/2mv2
✔️Now use kinematic equation again...
v2=u2+2gh
v 2 = 2gh
So,
➖Substitute the value again...
KE=1/2m(2gh)
=mgh
✔️(T.E.)C = K.E + P.E. = mgh +0 = mgh ...(iii)
✔️(TE)A =(TE)B=(TE)C
Hence, the total energy of the body is conserve during free fall..........
Attachments:
NidhraNair:
let me know in case of doubts...
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