Question

If we shift a body in equilibrium from $A$ to $C$ in a gravitational field via path $A C$ or $A B C$,

• A. The work done by the force $\vec{F}$ for both paths will be same
• B. $W_{A C} > W_{A B C}$
• C. $W_{A C} < W_{A B C}$
• D. None of the above

Answer 1

Answer: (A)

For the path $A C$,
$W_{A C}=F s \cos \left(90^{\circ}-\theta\right)=m g s \sin \theta=m g h$ $(\because F=m g)$
For path $A B, W_{A B}=F a \cos 90^{\circ}=0$
For path $B C, W_{B C}=F h \cos 0^{\circ}=m g h$
So that $W_{A B}+W_{B C}=m g h=W_{A C}$
i.e., $W_{A B C}=W_{B C}$