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Find relation between potential energy and kinetic energy at height x.【graph}】 ☑️Refer to attachment ⭕️ Help me in this.....Thanks☺️

Asked by jhajuhi19 | 05 Jul, 2019, 08:51: AM

Expert Answer

Potential energy PE of a particle undergoing projectile motion as a function of height h is given by, PE = m×g×h .........................(1)

where m is mass, g is acceleration due to gravity and h is height above projection point.

Kinetic energy KE of a particle undergoing projectile motion as a function of height h is given by, KE = (1/2)m×u² - (m×g×h) .................(2)

It can be seen from eqn.(1) and (2), PE and KE are linear function of h .

PE strats from 0 at projection point and increases at constant rate to maximum value at maximum height.

KE strats from maximum value at projection point and decreases at constant rate to zero at maximum height.

None of the graphs given in the question shows the above mentioned nature of variation for PE and KE.

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Normally, variable X is used for horizontal distance, not for height.

Hence I guess, intention of the question is to get PE and KE as a function of horizontal distance X but not of vertical distance h .

If we assume, it is required to get PE and KE as a function of horizontal distance X, the answer is explained below.

Potential energy PE = m×g×h = m×g×( u sinθ × t - (1/2)×g×t² ) .......................(3)

In eqn.(3), we used the relation for vertical distance h travelled for projectile motion. u is initial projection velocity,

θ is angle of projection and t is time when the particle reaches the height h.

Horizontal distance x travelled by the ball, X = ( u cosθ ) t .................(4)

If we eliminate time t in eqn.(3), using eqn.(4), we get , PE

begin mathsize 14px style equals space m space g space open parentheses X space tan theta space minus space g over 2 space fraction numerator X squared over denominator u squared cos squared theta end fraction close parentheses end style

.......................(5)

Eqn.(5) is inverted parabola about vertical axis , if X is chosen as horizontal axis.

Hence Graph(3) shows this behaviour for potential energy

Kinetic energy KE = (1/2) m×u² - m×g×h =

begin mathsize 14px style equals space 1 half m u squared space minus space m space g space open parentheses X space tan theta space plus space g over 2 space fraction numerator X squared over denominator u squared cos squared theta end fraction close parentheses end style

...........................(6)

Eqn.(6) is parabola about vertical axis , if X is chosen as horizontal axis.

Hence Graph(2) shows this behaviour for kinetic energy

Answered by Thiyagarajan K | 05 Jul, 2019, 03:14: PM

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