Question

Two change particles having $+ve$ charge are held together at a certain distance. When they are released they start moving away from each other and gain $150\text{J}$ of their $KE$. Determine the total work done by internal forces.

• A. $+150\text{J}$
• B. Zero
• C. None
• D. $-150\text{J}$

Answer 1

The correct option is A $+150\text{J}$


Since they are under force of repulsion (internal forces) and displacement is given in same direction.
Therefore work done will be $+ve$ and its magnitude will be the gain in their kinetic energy.

$\therefore$ Work done by internal force $=$ change in the particles $K.E$
$W=+150\text{J}$,
Here we get $+ve$ sign, the direction of force of attraction is same to the direction of displacement.

Hence option B is the correct answer