Class 11PhysicsWork, Energy, Power and CollisionWork Energy Theorem and Conservation of Mechanical Energy
Medium$A$ body falls freely from a height $h$ in a gravitational field. Match the following:
Column-$I$ Column-$II$ $(1)$ Potential Energy $=$ Kinetic Energy $(a)$ At height $\frac{2h}{3}$ $(2)$ Potential Energy $= 2 \times$ Kinetic Energy $(b)$ Constant at every point $(c)$ At height $\frac{h}{2}$
| Column-$I$ | Column-$II$ |
| $(1)$ Potential Energy $=$ Kinetic Energy | $(a)$ At height $\frac{2h}{3}$ |
| $(2)$ Potential Energy $= 2 \times$ Kinetic Energy | $(b)$ Constant at every point |
| $(c)$ At height $\frac{h}{2}$ |
(A) Let the total energy be $E = mgh$. At any height $y$ from the ground,Potential Energy $U = mgy$ and Kinetic Energy $K = mg(h-y)$.
For $(1)$,$U = K \implies mgy = mg(h-y) \implies y = h-y \implies 2y = h \implies y = \frac{h}{2}$. Thus,$(1) \rightarrow (c)$.
For $(2)$,$U = 2K \implies mgy = 2mg(h-y) \implies y = 2h - 2y \implies 3y = 2h \implies y = \frac{2h}{3}$. Thus,$(2) \rightarrow (a)$.
Therefore,the correct matching is $(1)-(c)$ and $(2)-(a)$.
For $(1)$,$U = K \implies mgy = mg(h-y) \implies y = h-y \implies 2y = h \implies y = \frac{h}{2}$. Thus,$(1) \rightarrow (c)$.
For $(2)$,$U = 2K \implies mgy = 2mg(h-y) \implies y = 2h - 2y \implies 3y = 2h \implies y = \frac{2h}{3}$. Thus,$(2) \rightarrow (a)$.
Therefore,the correct matching is $(1)-(c)$ and $(2)-(a)$.
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