Newton's Second Law

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Warm-up:

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Describing Newton 2:

... in words ...

$$\text{resultant force}=\text{mass}\times \text{acceleration}$$

... or in algebra ...

$$F=ma$$

• where
• $F$ is the resultant force on the object, in newtons, $N$;
• $m$ is the mass of the object, in kilograms, $kg$;
• and
• $a$ is the acceleration of the object, m/s².

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The Rearrangements:

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Questions:

F = ma Problems
(Care of stickmanphysics.com)

Newton's Second Law of Motion
(Care of www.docbrown.info)

Practice Problem Set
(Care of www.mrpowellscience.com)

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The Equation above is only useful if the mass of the object remains constant. In situations where this is not the case then use the following:

$$\mathbf{F}=\frac{d\mathbf{p}}{dt}$$

This can be derived from the first equation thus:

$$F=ma$$

Using the definition of acceleration we get:

$$F=m\frac{v-u}{t}$$ $$F=\frac{mv-mu}{t}$$

Using the definition of momentum we get:

$$F=\frac{p_f-p_i}{t}$$

where $p_i$ and $p_f$ are the momenta intially and finally respectively.

The change in momentum is $\Delta p=p_f-p_i$, so,

$$F=\frac{\Delta p}{\Delta t}$$

This form is useful either when the rate of change is constant over time interval (or near constant) where the rate is changing rapidly then we take the limit and get,

$$\mathbf{F}=\frac{d\mathbf{p}}{dt}$$

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Kinetic Energy

Prerequisite Materials:

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Describing Kinetic Energy

Word equation ...

$$\text{Kinetic Energy}=\frac{1}{2}\times \text{mass}\times {\text{velocity}}^2$$

The algebra ...

$$E_{\text{Kinetic}}=\frac{1}{2}m{v}^2$$

• where
• $E_{\text{Kinetic}}$ is the kinetic energy of the object in joules, $J$
• $m$ is the mass of the moving object on kilograms, $kg$
• and
• $v$ is the speed, the magnitude of the velocity, of the object in metres per second, $m{s}^{-1}$

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The Rearrangements

$$m=\frac{2E_{\text{Kinetic}}}{v^2}$$ $$v=\sqrt{\frac{2E_{\text{Kinetic}}}{m}}$$

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Questions:

Worksheet
(Care of gcserevisionbuddy.co.uk)

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... in Terms of Momentum

$$E_{\text{Kinetic}}=\frac{1}{2}m{v}^2$$

Multiplying the RHS by $\frac{m}{m}$

$$=\frac{m^2{v}^2}{2m}$$ $$=\frac{p^2}{2m}$$

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Further study:

Potential Energy

Work Energy Principle

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