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

Options

A. (a) The work done by the force `vecF` for both paths will be same
B. (b) `W_(AC)gtW_(ABC)`
C. (c) `W_(AC)ltW_(ABC)`
D. (d) None of the above

Answer

A

Explanation

For the path AC,
`W_(AC)=Fs cos (90^@-theta)=mg sin theta=mgh` ( `:'F=mg`)
For path AB, `W_(AB)=Fa cos 90^@=0`
For path BC, `W_(BC)=Fhcos0^@=mgh`
So that `W_(AB)+W_(BC)=mgh=W_(AC)`
i.e., `W_(ABC)=W_(BC)`