# Help to understand the concept of Power

Power is described as the Velocity at which Force is applied to an object (P=F*d/t)

However, how are the Force and Velocity independent? Doesn't the Force applied affect the Velocity of the object?

For example, let's say you want to lift a 19.62N (2kg*9.81m/s^2) object upwards 1 meter in 1 second. The way I see the problem solved online is that you would need 19.62W = (19.62N*1m/s). However, with this applied Force the object is in equilibrium (and not moving or moving at an arbitrary constant velocity), so the work and power would be zero.

I must have something wrong here. Thanks.

The object gains potential energy as it rises. Gravity can be considered a fictitious force.

A.T.
moving at an arbitrary constant velocity), so the work and power would be zero.
If it's moving at constant velocity, then the work and power aren't zero.

Nidum
Gold Member
Power is described as the Velocity at which Force is applied to an object (P=F*d/t)

That's a very bad definition .

There are formal definitions of Power but for practical purposes :

Power is rate of doing work .
or
Power is rate of transfer of energy from a source to a sink .

Last edited:
Drakkith
Staff Emeritus
Power is described as the Velocity at which Force is applied to an object (P=F*d/t)

However, how are the Force and Velocity independent? Doesn't the Force applied affect the Velocity of the object?

Yes, the force would affect the velocity of the object, but it doesn't matter. That fact is already "built-in" to work and power. For example, applying a net force of 10 newtons for 10 seconds on a 1kg object:

##a = \frac{f}{m} = \frac{10}{1} = 10 m/{s^2}##
##v = v_0 + at = 0 + 10*10 = 100 m/s##

The distance the object moves over this time period is given by:
##x = x_0 + v_0t + 1/2at^2 = 0 + 0 + 1/2*10*10^2 = 500 m##

The work required to get the object up to 100 m/s is just the same as its kinetic energy.
##k_e = 1/2mv^2 = 1/2*1*100^2 = 5,000 J##

We can also calculate the work done using the force and the distance:
##W = f*d = 10 * 500 = 5,000 J##

Power:
##P = W/t = 5,000/10 = 500 W##

For example, let's say you want to lift a 19.62N (2kg*9.81m/s^2) object upwards 1 meter in 1 second. The way I see the problem solved online is that you would need 19.62W = (19.62N*1m/s). However, with this applied Force the object is in equilibrium (and not moving or moving at an arbitrary constant velocity), so the work and power would be zero.

You need to use the net force on the object in your calculation. In equilibrium there is no net force and no work is performed on the object by any of the forces. Note that you will still expend energy and perform work on yourself because your body is not a perfectly efficient machine. Your muscles fibers constantly contract and relax just to hold an object at rest, your bones and muscles stretch and bend, the chemical reactions powering your muscles release a lot of energy simply as heat, etc.