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Kinetic energy is defined in my textbook to be the work required to reduce an object's speed to zero.

If a object is moving with some speed and has a mass then we need some force to brake it. Is the total amount of work (not force) needed the same as the kinetic energy it possesses?

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  • $\begingroup$ No. There is also the distance the object travelled as it braked. Like a car. So you need to multiply force to distance $\endgroup$ – QuIcKmAtHs Jan 6 '18 at 8:18
  • $\begingroup$ @XcoderX right, amouting to "work". $\endgroup$ – user1 Jan 6 '18 at 8:28
  • $\begingroup$ @XcoderX ah I wrote force! $\endgroup$ – user1 Jan 6 '18 at 8:29
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    $\begingroup$ The braking force is the braking force... You need to multiply distance $\endgroup$ – QuIcKmAtHs Jan 6 '18 at 8:30
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A force is not an energy so you can't compare a force and an energy and it doesn't make sense to ask if the total amount of force needed the same as the kinetic energy?

However force times distance is an energy, and indeed it is the definition of work i.e. work = force $\times$ distance. So what we can say is that the work is the same as the change in kinetic energy. In the simple case where the force is constant we get:

$$ F s = \Delta KE $$

where $F$ is the force, $s$ is the distance travelled and $\Delta KE$ is the change in the kinetic energy. If the force is variable then we would have to use an integral:

$$ \int_0^s F(s') ds' = \Delta KE $$

One final complication: the equation above only applies if the force is applied in the direction of motion. If the force isn't in the direction of motion then we have to treat the force and distance moved as vectors and our equation becomes:

$$ \int_0^s \mathbf F(\mathbf s') \cdot d\mathbf s' = \Delta KE $$

where the dot means the dot product.

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  • $\begingroup$ So say the kinetic enegry is $1$ J then we need a force and a distance multiplying to that number to reduce the speed to $0$? $\endgroup$ – user1 Jan 6 '18 at 8:43
  • $\begingroup$ @user1 yes e.g. a force of $1$N applied for $1$m, or $10$N applied for $0.1$m, and so on $\endgroup$ – John Rennie Jan 6 '18 at 8:46
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The equation would be $\frac{1}{2}mv^2 = Fd$.

Here, the kinetic energy would be the constant braking force, multiplied by the distance the object traveled.

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