Since energy can be neither created nor destroyed, external work performed on a system changes the system’s total energy.
$latex W = {\Delta}E = E_2 – E_1$
The work energy principle introduces some simplifications into many mechanical problems, such as:
- You don’t need to know the acceleration of a body to calculate the work performed on it.
- Forces normal to the direction of motion (i.e. don’t contribute to work) are not considered.
- You don’t need to work in vector quantities. Only scalars are involved.
- You can analyze the system as a whole, not in individual parts.
Example
An elevator starts from rest, accelerates uniformly to a constant speed of 2.5 m/s, and then decelerates uniformly to a stop 100 m higher than its initial position. The elevator weighs 3000 kg. What work was done on the elevator?
Solution
$latex W = E_2 – E_1\newline
\-\hspace{7mm}= mg(h_2 – h_1)\newline
\-\hspace{7mm}=(3000 kg)(9.81 m/s^{2})(100 m) = 2,943 kJ$
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