Question
What is a conservative force? Prove that gravitational force is conservative, while frictional force is non-conservative.
✓
Answer
Conservative force : The work done against or by the force in moving a body depends only on initial and final positions of body and not on nature of path followed. For example, gravitational force is a conservative force.
Consider a body of mass 'm' to be moved to height ‘h’ over different paths from x to y.
In Mass is being raised vertically upwards along direction of force applied.
Work done = Force x distance
W1 = mg x h
In Mass 'm' is taken along a smooth inclined plane yz of height 'h' and inclination a.
Work done = F × yz
W2 = mg sin a × yz
$=\text{mg}\Big(\frac{\text{xy}}{\text{yz}}\Big)\times\text{yz}$
(F = mg sina as sine component contributes to applied force)
W2 = mgh
In Mass ‘m' is taken to height 'h' through a staircase having ‘n’ number of steps. Each step has height h' and width x' so work done in carrying the mass through horizontal distance $\text{x}'=\text{mg}\cos90^\circ=0$ and through vertical distance h' = mgh'.
Hence total work done in n steps is W3 = mgh' × n = mgh
In Mass ‘m’is carried to height-h’ in a zig-zag path. Zig-zag path can be divided into infinitesimally small steps of horizontal displacement dx and vertical displacement dh. As the work done along any horizontal displacement is zero. Total work done along all vertical paths will add up $\text{W}_4=\Sigma(\text{dh)}=\text{mgh}$
Thus $\text{W}_1=\text{W}_2=\text{W}_3=\text{W}_4=\text{mgh}$
$\therefore$ Whatever is the path followed to go from X to Y work done is the same. This shows that gravitational force is a conservative force.
Consider a body of mass 'm' to be moved to height ‘h’ over different paths from x to y.
In Mass is being raised vertically upwards along direction of force applied.
Work done = Force x distance
W1 = mg x h
In Mass 'm' is taken along a smooth inclined plane yz of height 'h' and inclination a.
Work done = F × yz
W2 = mg sin a × yz
$=\text{mg}\Big(\frac{\text{xy}}{\text{yz}}\Big)\times\text{yz}$
(F = mg sina as sine component contributes to applied force)
W2 = mgh
In Mass ‘m' is taken to height 'h' through a staircase having ‘n’ number of steps. Each step has height h' and width x' so work done in carrying the mass through horizontal distance $\text{x}'=\text{mg}\cos90^\circ=0$ and through vertical distance h' = mgh'.
Hence total work done in n steps is W3 = mgh' × n = mgh
In Mass ‘m’is carried to height-h’ in a zig-zag path. Zig-zag path can be divided into infinitesimally small steps of horizontal displacement dx and vertical displacement dh. As the work done along any horizontal displacement is zero. Total work done along all vertical paths will add up $\text{W}_4=\Sigma(\text{dh)}=\text{mgh}$
Thus $\text{W}_1=\text{W}_2=\text{W}_3=\text{W}_4=\text{mgh}$
$\therefore$ Whatever is the path followed to go from X to Y work done is the same. This shows that gravitational force is a conservative force.
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