The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
The answer to this question is a) sulfur
The engine's efficiency is (35J)/125J) = 28% .
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