# Work done against a force

Consider a body of $1\ \rm kg$ mass at rest at ground. Now as we know, it can be lifted by a force equal to its weight. So we have to apply a little bit more force then its weight. let $g= 10\ \rm ms^{-2}$, its weight is $10\ \rm N$. So I apply a force of $20\ \rm N$ or $10\ \rm N$ net force. As we know all the work I shall do will be converted to P.E of the mass. In case of friction my work done is wasted as heat.But what in the case if we consider a frictionless world for a while and there is no gravity now if we consider a mass moving with constant velocity in space with $10\ \rm N$ force applied on it and an opposing force of $-10\ \rm N$ acting on it where the work done by $10\ \rm N$ force against $-10\ \rm N$ will go?

• It will go into the object, which supplies the 10 N. That pulling force must come from somewhere - maybe a rocket (so the energy is spent to burn and accelerate fuel) or a gravitational field (so the energy is spent "moving" the planet / stored as potential energy) or alike – Steeven Aug 7 '18 at 13:12
• Just to be clear, are you saying there are no forces (gravity, friction, etc.) acting on the mass except for 2 equal and opposite forces and the velocity is constant? – Bob D Aug 7 '18 at 13:28
• Of curse! I an talking about the same thing – AHTSHAM KHALID Aug 25 '18 at 13:28
• Please, separate your question in paragraphs, and use commas and full stops. If you make reading easy, you'll surely get better and more useful answers. Try to state your question in a calm and organised way. – FGSUZ Sep 1 '18 at 22:52

## 2 Answers

The formula for work is:

The work $W$ done by a constant force of magnitude $F$on a point that moves a displacement $s$ in a straight line in the direction of the force is the product $W = F \cdot s$.

When we are talking about the world without friction, and the body (point) is moving straight with constant velocity we can say in accordance with the Newton's first law that:

In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.

In your case where two equal forces $\vec{F_1} = -\vec{F_2}$ act on a point in opposite directions, in accordance with the superposition principle: $\vec{F_1} +\vec{F_2} = 0$.

Then based on $\vec{F} = m\vec{a}$, follows $\vec{a} = 0$ and velocity is a constant, moreover by inertial frame introduction above we can set $s = 0$, in inertial frame, where point is at a rest.

If we talk about $F_1$, and we want to calculate the work $W_1$ done by $F_1$, using the formula above we get $W_1 = F_1 \cdot 0 = 0$.

• Just because net force is $0$ does not mean displacement is $0$. – Aaron Stevens Sep 1 '18 at 22:08
• Corrected an answer about inertial frame – Artem Sep 1 '18 at 22:12
• "if we consider a mass moving with constant velocity". The OP is considering a scenario where we are in an inertial frame and we observe the object moving at constant velocity with two equal but opposite forces acting on it. I can see what you are doing here now, but you should explicitly tell the OP in your answer why this new scenario you have created still answers the original question. – Aaron Stevens Sep 1 '18 at 22:21

In the case of an object moving at a constant velocity with two equal but opposite forces acting on the object, the net work done on the object is $W=\int \vec F_{net} \cdot d\vec s= \int \left (\vec F_1-\vec F_2\right ) \cdot d\vec s=0$. One force inputs energy and the other removes energy.

Without knowing more about the forces in question we cannot say where the energy is coming from or going to. For example if we are dealing with two conservative forces then the potential energy from one force is being converted into potential energy of the other force. If we allow friction back into our world and one of the forces is non-conservative (like some dissipative force), then the energy will be lost to heat.

All we can say is that the object will not be changing its kinetic energy.

• OP indicated that its a "frictionless world" so all forces are conservative. – Artem Sep 1 '18 at 22:47
• @Artem Thanks. I have addressed this in my answer. – Aaron Stevens Sep 1 '18 at 22:51