**Kinetic Energy and Movement (Velocity).-
To obtain the relation between energy and movement, let's imagine a particle
of mass m moving in a straight line with initial velocity V _{i }.
A net constant force F is then applied parallel to its movement, over a distance
d. Then, the work done over the particle is W = Fd.
As F = ma (a, acceleration) and using the kinetic energy formula V_{f}^{2}
= V_{i}^{2} + 2ad, where V_{f} is the final velocity,
we get:**

**W = Fd = mad = m[(V _{f}^{2} - V_{i}^{2})
/ 2d]d**

**That is, W = ½mV _{f}^{2} - ½mV_{i}^{2}
**

**It is clear we have a difference between
final and initial quantities.**

** The kinetic energy (translational energy)
of a particle is defined by physicists as the quantity ½mv ^{2} .**

**Ec = ½mv ^{2}.**

**W can also be written**

**W =
Ec**

**That is, the net work done on an object
is equal to the change in his kinetic energy. This result is known
as the work-kinetic energy theorem.**

**Let's notice W is the net work done
over the object.**

**Example. Starting from rest, you push
your 1.000 kg car over a 5 meters distance, on an horizontal ground,
applying an also horizontal 400 N force. What is the car kinetic energy change?;
What is its final velocity at the end of the 5 meters displacement?
Disregard any friction force.**

**Solution. The change in kinetic energy
must be equal to the net work done on the car, W = Fd**

**The final velocity is obtained from the equation
W = ½mV _{f}^{2} - ½mV_{i}^{2}, where V_{i} =
0.**

**2.000 J = ( ½ )(1000 kg)V _{f}^{2},
from where V_{f} = 2 m/s. **

**Physics Homework, Visit Sites:
· Energy, Work and Power: Concepts
· Kinetic Energy
· Potential Energy
· Power
**

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· Physics, Main Page
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· Exercises Using Coulomb's Law
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