S3.4 Kinetic energy
[A] The work - energy theorem
- The work - energy theorem defines kinetic energy.
Key Point 3.7
Work-energy theorem. When a mass m is accelerated by a force along some path, the total work done on the mass by the force is
![\[ W = \int^f_i \vec{F} \cdot d \vec{r} = K_f - K_i \]](mastermathpng-0.png)
where the initial and final kinetic energies are Ki and Kf respectively.
Analysis
Reveal
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Analysis
It is straightforward to derive the work-energy theorem from results you have met already.
Consider the situation where a constant magnitude of force acts on an object increasing its velocity from v1 to v2 while moving it over a distance s. Then from Key Point 1.4
![\[ v_2^2=v_1^2 + 2as \]](mastermathpng-3.png)
Rearranging gives
![\[ a=\frac{v_2^2-v_1^2}{2s} \]](mastermathpng-4.png)
and thus
![\[ F=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{m\frac{v_2^2-v_1^2}{2s}} \]](mastermathpng-5.png)
So that
![\[ Fs=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2} \]](mastermathpng-6.png)
Identifying
, we obtain Key Point 3.7
Example
Q: A golf ball is rolled in a straight line, say along the x-axis. Does its kinetic energy increase, decrease or stay the same if:
- [(a)] Its velocity changes from -4ms-1 to -1ms-1 ?
- [(b)] Its velocity changes from -4ms-1 to 4ms-1 ?
In each case is the work done positive negative or zero?
Solution
[(a)] From Key Point 3.7 and Key Point 3.8
![\[ W = K_f - K_i = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \]](mastermathpng-8.png)
Since
, the kinetic energy of the golf ball decreases and the work done is negative.[(b)] Here
, so no change in kinetic energy and from
Key Point 3.7, no work is done.Q. Why no work?
A. In changing the ball’s velocity from -4ms-1 to rest, negative work is done, while changing the velocity from rest to 4ms-1, the same amount of positive work is done.
Learning Resources
| HRW 7.5 | |