# The Block’s Acceleration

Worked Example 16 Newton’s Second Law
Question: A block of mass 10 kg is accelerating at 2 m·s−2. What is the magnitude of the net force acting on the block?

Step 1:

We are given

the block’s mass
the block’s acceleration
all in the correct units.

Step 2 :

We are asked to find the magnitude of the force applied to the block. Newton’s Second Law tells us the relationship between acceleration and force for one object. Since we are only asked for the magnitude we do not need to worry about the directions of the vectors:

{\displaystyle {\begin{matrix}F_{Net}&=&ma\&=&10\ {\mbox{kg}}\times 2{\mbox{ m}}\cdot {\mbox{s}}^{-2}\&=&20\ {\mbox{N}}\end{matrix}}} {\displaystyle {\begin{matrix}F_{Net}&=&ma\&=&10\ {\mbox{kg}}\times 2{\mbox{ m}}\cdot {\mbox{s}}^{-2}\&=&20\ {\mbox{N}}\end{matrix}}}
Thus, there must be a net force of 20 N acting on the box.

Worked Example 17 Newton’s Second Law 2
Question: A 12 N force is applied in the positive x-direction to a block of mass 100 mg resting on a frictionless flat surface. What is the resulting acceleration of the block?

Step 1 :

We are given

the block’s mass
the applied force
but the mass is not in the correct units.

Step 2 :

Let us begin by converting the mass:

{\displaystyle {\begin{matrix}100{\mbox{ mg}}&=&100\times 10^{-3}{\mbox{ g}}=0.1{\mbox{ g}}\1000{\mbox{ g}}&=&1\ {\mbox{kg}}\1&=&1kg\times {\frac {1}{1000g}}\&=&{\frac {1kg}{1000g}}\0.1g&=&0.1g\times 1\&=&0.1g\times {\frac {1kg}{1000g}}\&=&0.0001\ kg\\end{matrix}}} {\displaystyle {\begin{matrix}100{\mbox{ mg}}&=&100\times 10^{-3}{\mbox{ hich {\displaystyle 2\times 2} {\displaystyle 2\times 2} is only 2 gatherings of 2 included to give 4. We cone embrace a comparative way to deal with see how vector duplication functions.

Fhsst vectors32.png

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Vectors

PGCE Comments – TO DO LIST – Introduction – Examples – Mathematical Properties – Addition – Components – Importance – Important Quantities, Equations, and Concepts

Presently which you have been familiar without the numerical properties of vectors, we come back to vector expansion in more detail. There are various procedures of vector expansion. These procedures fall into two principle classes graphical and arithmetical systems.

Graphical Techniques

Graphical methods include attracting precise scale outlines to signify singular vectors and their resultants. We next talk about the two essential graphical procedures, the tail-to-head strategy and the parallelogram technique.

In portraying the scientific properties of vectors we utilized removals and the tail-to-head graphical strategy for vector expansion as one outline. In the tail-to-head technique for vector expansion the accompanying system is pursued:

Pick a scale and incorporate a reference bearing.

Pick any of the vectors to be summed and draw it as one bolt in the right bearing and of the right length-make sure to put one pointed stone on the conclusion to signify its course.

Take the following vector and draw it as one bolt beginning from the sharpened stone of the principal vector in the right course and of the right length.

Proceed until you have drawn every vector-each time beginning from the leader of the past vector. Along these lines, the vectors to be included are drawn consistently tail-to-head.

The resultant is then the vector drawn from the tail of the principal vector to the leader of the last. Its extent cone be resolved from the length of its bolt utilizing the scale. Its course also cone be resolved from the scale chart.