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PART - 1 CH - 5Work, Energy and Power — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 5Work, Energy and Power3 Marks
Question
Prove that as the object moves towards the ground from some height, its potential energy decreases and kinetic energy increases by the same amount.
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Answer

 Suppose an object of mass $m$ is at rest at point A at a height $h$ from the earth's surface as shown in figure. At point A, object will have only potential energy and its kinetic energy will be zero. Value of potential energy at point $A = mgh$. Suppose the object falls freely and when the object reaches the ground, its speed is $v$ hence
the value of kinetic energy at point B will be $\frac{1}{2} m v^2$. Here, $h$ $=0$ hence potential energy $U =0$.

Image
From the third equation of motion the speed of object at point $B$ is
$\begin{aligned}
v^2 & =u^2+2 a s \\
v^2 & =0+2 g h \\
\therefore \quad v^2 & =2 g h
\end{aligned}$
$\therefore \quad$ Kinetic energy at point B
$\begin{array}{l}
K=\frac{1}{2} m v^2 \\
K=\frac{1}{2} m \times 2 g h=m g h \\
K=m g h
\end{array}$
That is, the value of kinetic energy of an object on earth's surface is equal to the potential energy at point A . The decrease in potential energy of a freely following object is equal to increase in its kinetic energy. As the objects comes from height towards the ground the potential energy decreases and the kinetic energy increases by the same amount.
$\Delta K=-\Delta U$

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