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Q5.7(c):
State if the following statement is true or false. Give reasons for your answer. Work done in the motion of a body over a closed loop is zero for every force in nature.

Solution :

Statement Assessment

The statement "Work done in the motion of a body over a closed loop is zero for every force in nature" is False.

Theoretical Justification

Work done ($W$) in moving a particle along a closed path is defined by the line integral of the force vector ($\vec{F}$) over the displacement vector ($d\vec{r}$):

$W = \oint \vec{F} \cdot d\vec{r}$

The validity of this integral being zero depends entirely on the nature of the force field acting on the body. Forces are categorized into two distinct classes based on their mathematical properties:

  • Conservative Forces: These forces possess the property that the work done depends only on the initial and final positions, not the path taken. By definition, for any conservative force, the work done over a closed loop is identically zero ($\oint \vec{F}_c \cdot d\vec{r} = 0$). Examples include gravitational force, electrostatic force, and the spring force. [Per the Conservative Field Theorem].
  • Non-Conservative Forces: These forces do not satisfy the path-independence condition. In the presence of non-conservative forces, the work done along a closed loop is non-zero, as energy is typically dissipated (usually converted into heat or sound) over the duration of the cycle. Examples include kinetic friction, air resistance, and viscous drag. [Per the Work-Energy Theorem].

Mathematical Counter-Example: Kinetic Friction

Consider a block sliding on a rough horizontal surface along a closed rectangular path of side lengths $L$ and $W$.

The frictional force ($f_k$) always acts in the direction exactly opposite to the instantaneous velocity ($v$). Therefore, the angle between $\vec{F}$ and $d\vec{r}$ is always $180^\circ$ ($\pi$ radians).

Using the definition of work ($W = \int |\vec{F}| |d\vec{r}| \cos \theta$):

$W = \oint f_k \, ds \, \cos(180^\circ) = -f_k \oint ds$

$W = -f_k \times (\text{Total Perimeter}) = -f_k \times (2L + 2W)$

Since $f_k > 0$, the total work done $W \neq 0$.

Visual Representation of a Closed Path under Non-Conservative Force

Velocity ($v$) Friction ($f_k$)

Summary Analysis

Force Type Work over Closed Loop ($\oint \vec{F} \cdot d\vec{r}$)
Conservative (Gravity, Electrostatic) $0$
Non-Conservative (Friction, Viscosity) $\neq 0$

Final Solution: The statement is False. Work done in a closed loop is zero only for conservative forces. For non-conservative forces, such as friction or air resistance, the work done over a closed loop is non-zero because the energy is dissipated into the surroundings.


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