Derivation of Potential Energy {P.E.} class 9th
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Answer:
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Derivation
Simply we know that, work done is product of force & displacement.
→ W = F × s
In case of gravitational potential energy (according to Class 9th), force which stands for m × a, where a will be replaced by g (gravitational acceleration) & s (displacement) by h (height).
→ P.E. = mg × h
→ P.E. = mgh
Extra Knowledge:
Do energy get conserved? Proof
Case 1: using K.E. formula ½ mv².
→ K.E. = ½ × mv²
If object is at rest then, v = 0.
→ K.E. = ½ × m × 0² = 0 (zero)
→ P.E = mgh (it will remain as it's formula because currently object is at rest)
Total Energy = mgh + 0 = mgh
Case 2: object comes into motion, applying v² = u² + 2as
We know that object was at rest, hence u = 0.
→ v² = 0² + 2gh [here a = g, s = h]
→ v² = 2gh __(i)
→ K.E = ½ × m × 2gh
→ K.E. = mgh
Potential Energy turns 0 (zero).
Total Energy = mgh + 0 = mgh
Case 3: when object has covered almost k distance out of h meters.
→ P.E. = mg(h - k)
→ K.E = ½ × mv²
In eq(i), it was h but since now it has covered k distance hence h will be replaced by k.
→ K.E. = ½ × m × 2gk
→ K.E. = mgk
Total Energy = mg(h - k) + mgk
→ mgh - mgk + mgk
→ mgh
Hence we find total Energy to be conserved.