# I am having trouble understanding work and the factor of time

Consider this:

• 1 kg object vs 1000 kg object
• No friction applies in my example and the machine I am using is 100% efficient.
• Some machine using a fuel applies 1 N of force over 1 meter to each object which equates to 1 J of transferred energy.
• It takes 1.41 s for the 1 kg object to reach 1 meter. It takes about 45 s for the 1000 kg object to reach the same distance.
• Applying 1 N for 45 s surely must require more energy input to the machine than applying 1 N for only 1.41 s. Yet both objects acquire only 1 J (due to the kinetic energy giving more importance to velocity than mass) as the 1000 kg object gain only a fraction of the velocity that the 1 kg object does.

But seeing that 1 N of force throughout 45 s CLEARLY requires more joules than 1 N throughout 1.41 s, where does the rest of the joules go? I mean, energy is a conservative thing, right?

Where does my reasoning fail?

• "CLEARLY" I would say clearly not. You use 1 joule, there is no "rest of the energy". Energy is not "conservative". – my2cts Jan 7 at 19:03

It doesn't matter how long it takes for an object to move some distance due to an applied force. The definition of work is just $$W=\int\mathbf F\cdot\text d\mathbf x$$
This definition says nothing about how long it takes for the force to be applied over the displacement. So in a 1D example, a force of $$1\ \rm N$$ applied over a distance of $$1\ \rm m$$ corresponds to $$1\ \rm J$$ of work if it takes $$1\ \rm s$$ to complete or $$1000$$ years.
In this case you do get something for your extra effort, but it is not energy. When the pushing is done the kinetic energy $$1/2\ m\ v^2$$ is the same for both objects, but the heavier object has more momentum in the same ratio ~$$45/1.4$$ since $$\Delta (m v)=F \Delta t$$.