A body of mass $5 k g$ falls from a height of $10 m$ to $4 m$ . Calculate: (i) the loss in potential energy of the body. (ii) the total energy possessed by the body at any instant? (Take $g = 10 m s^{- 2}$ )

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Solution

(i) Mass of the body $= 5 k g$ $g = 10 m / s^{2}$
P.E. at height $10 m = m g h = 5 \times 10 \times 10 = 500 J$
P.E. at height $4 m = m g h = 5 \times 10 \times 4 = 200 J$
Loss in P.E. $= (500 - 200) J = 300 J$
(ii) The total energy possessed by the body at
Any instant remains constant for free fall
It is equal to the sum of P.E. and K.E.
$∴$ At height $10 m$ , i.e.. at top most point. $K . E . = 0$
$∴$ Total energy $= (P E_{T o p} + K E_{T o p})$
Total energy $= 500 + 0 = 500 J$

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Similar questions

A body of mass 5 kg falls from a height of 10 m to 4m. Calculate:

(i) the loss in potential energy of the body, and (ii) the total energy possessed by the body at any instant? (Take g=10 m $s^{- 2}$ ).

5 k g

10 m

4 m

g = 10 m s^{- 2}

5 k g

10 m

4 m

g = 10 m s^{- 2}

g = 10 m s^{- 2}

4 k g

10 m

g = 10 m s^{- 2}

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