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I'm reading mechanics at undergraduate level and I'm wondering if I understand power correctly.

Let's suppose we have mass and we push on it with the constant net force $ \mathbf{F} $ along the positive x-axis (there is no friction.) If $ \mathbf{v_{t_1}}$ is the velocity along x-axis at some time $t_1$ the power is $ \mathbf{F} \cdot \mathbf{v_{t_1}} $

This implies that the power increases with increasing velocity. So after some time $t_2 = t_1 + dt$ it is true that $ \mathbf{F} \cdot \mathbf{v_{t_1}} < \mathbf{F} \cdot \mathbf{v_{t_2}} $

Is this correct?

If so it just feels strange to me as I have earlier thought, in high school, that the power is constant if we apply a constant force on a mass. One of the reasons I think it's strange is that if the velocity of the mass goes to light-speed then a very small push on the mass will change the power dramatically and will cause a big change in the kinetic energy of the mass. Where as if you push, with the same force, on the mass when it has a low velocity it will cause a small change in power and kinetic energy.

If all this is correct then I think it implies that if we want to dramatically change the kinetic energy of an object, we should apply a force on it when the object has a very large velocity. (I don't know how this would turn out in practice though.)

Kyle Kanos
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SwedeGustaf
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2 Answers2

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Power is, of course, the rate of energy transfer.   The amount of energy transferred is the product of force and distance.   In fact, $P=\mathbf{F\cdot{v}}$   is actually a rewriting of   $$P={\mathbf{F\cdot\frac{d}{\mathsf{t}}}}$$

In the absence of friction, the mass is accelerated by the constant force.   Although the change in velocity, $a\Delta t$, at a later time interval is the same as the change in velocity, $a\Delta t$, at an earlier time interval of the same duration, the distance traveled during the later time interval, $$\mathbf{d_2}=\mathbf{v_{t_2}}\Delta t+a\Delta t$$ is greater than $$\mathbf{d_1}=\mathbf{v_{t_1}}\Delta t+a\Delta t$$ So yes, the power is increasing with increasing velocity.

At speeds growing significantly close to light speed, it can't be ignored that $a$ gets smaller and smaller with the same constant applied force, because the inertial mass gets bigger and bigger. So the power would approach a constant value.

D. Ennis
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$\text{work done = force}\times \text{displacement}$

$\text{power} = \dfrac{\text{work done}}{\text{time taken}} = \text{force} \times \text{velocity}$

So if a force moves a greater distance in unit time the rate of working (power) will be larger.

Drop a mass of $2 \,\rm kg$ at rest and allow it to fall $5\, \rm m$.

Instantaneously at the start the power will be zero and at the end it will be $200 \, \rm W$ the fall taking a time of one second and the loss of gravitational potential energy (and gain in kinetic energy) being $100 \, \rm J$.

Now look at the motion over the next second during which the mass falls a distance of $15\, \rm m$.

At the start the power will be $200 \, \rm W$ and at the end it will be $400 \, \rm W$ the fall taking a time of one second and the loss of gravitational potential energy (and gain in kinetic energy) being $300 \, \rm J$.

Possible your uncertainty is due the fact that one quantity is dependent on distance and the other dependent on time?
So trelling further in a given time requires the expenditure of more energy.

Farcher
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