Question

A particle is taken from point A to point B under the influence of a force field. Now it is taken back from B to A and it is observed that the work done in taking the particle from A to B is not equal to the work done in taking it from B to A. If $W_{nc}$ and $W_c$ are work done by non-conservative and conservative forces present in the system, respectively, $\Delta U$ is the change in potential energy and $\Delta K$ is the change in kinetic energy, then

• A. $W_{nc}-\Delta U=\Delta K$
• B. $W_c=-\Delta U$
• C. $W_{nc}+W_c=\Delta K$
• D. $W_{nc}-\Delta U=-\Delta K$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $W_{nc}-\Delta U=\Delta K$
B $W_c=-\Delta U$
C $W_{nc}+W_c=\Delta K$

Work done by all forces is the change in kinetic energy.

So,

$W_c+W_{nc}=\Delta K⟶\left(I\right)$

Also,

Only the conservation force can change the potential energy of the system. So,

$W_c=-\Delta 4⟶\left(II\right)$

From equations $\left(I\right)$ & $\left(II\right)$ we get,

$W_{nc}-\Delta 4=\Delta K$