Newton’s second law of motion

We have seen in a previous post the Newton’s first law of motion.

Today we are going to see the second law of motion.

From our discussion in the first law of motion and in the post on acceleration when a force acts on an object either velocity increases due to a change in speed or the velocity changes due to a change in direction. Remember velocity is a vector quantity hence it will change if either the direction or the magnitude changes.

Hence we can deduce that if a force acts on an object then the velocity of the object changes.The more force is applied the greater the change in velocity. This must mean that if the mass remain constant then the momentum of the object changes. And hence the more force is applied the more the linear momentum will change.

Let us look at the situation below:

A force F acts on an object of mass m for a time t such that before the application of the force the object has a velocity v₁ hence a linear momentum of p₁ and after the force is applied the velocity of the object is v₂ hence the linear momentum is p₂.

Before the force is applied velocity = v₁ and linear momentum = p₁

After the force is applied velocity =va₂  and linear momentum is p₂.

Now the Newton’s second law of motion states that the force applied is directly proportional to the rate of change of linear momentum of the object.

If = a then

If k = 1

Then this will reduce the equation to

F= ma

This where the famous equation F = ma come from. This why it is also considered as the Newton’s second law of motion

Second equation of motion

We have seen in a previous post the first equation of motion.

Today we are going to see the second equation of motion.

You would remember that the first equation dealt with acceleration, time taken, initial velocity and final velocity.

So how do we derive the second equation of motion?

Let us take an object that starts from an initial velocity u and then accelerates during a time t until it reaches a final velocity v.

Now during the time the object was accelerating the object has moved a distance s. In the second equation of motion we will try to derive the equation to calculate this distance s.

Fig 2 below shows what is happening to the object.

Fig 1

Now you would remember that the area under a velocity-time graph is the distance travelled by the object. Hence we are going to calculate the area under the graph as indicated by the shaded area below.

Fig 2

Now if s the distance travelled is the area under a velocity-time graph then we can say that

Area under graph = ¹/₂( v + u)*t

hence s = ¹/₂( v + u)*t

Now if you can remember the first equation of motion is

v = u +at

Hence if we replace v = u +at into s = ¹/₂( v + u)*t we get

s = ¹/₂( v + u)*t

s = ¹/₂( [u +at] + u)*t

s = ¹/₂( u + at + u)*t

s = ¹/₂(2 u + at )*t

s =¹/₂(2ut +at²)

S = ut + ¹/₂at²

Hence as you can see above the second equation of motion is

S = ut + ¹/₂at²

We are now going to look at two examples where the second equation of motion can be used.

Example 1

A bus starts from rest and accelerates at a rate of 2.5 m/s² for a time of 50 s. Determine the distance travelled by the bus during the acceleration phase.

Now let us identify the different variables involved.

u the initial speed = 0 m/s

t time taken = 50 s

a acceleration = 2.5 m/s²

s distance travelled = ???

If we want calculate s the distance travelled we will have to use the second equation of motion

S = ut + ¹/₂at²

Substituting the different values we get

S = ut + ¹/₂at²

= o*50 + 0.5*2.5 *50²

= 0 + 1.25*2500

=3125 m

Example 2

An aero plane travelling at a speed of 300 m/s lands on a track 2000 m long and decelerates for a time of 25 s until it comes to rest. Calculate the deceleration needed to stop the plane.

We must now identify the different variables involved as in the first example.

initial speed u = 300 m/s

time taken t = 25 s

Distance travelled s = 2000 m

acceleration a = ????

If we want calculate the acceleration a we will have to use the second equation of motion

S = ut + ¹/₂at²

and make a the subject of formula.

s – ut = ¹/₂at²

a =2(s - ut)/t²

= 2(2000 – 300*25)/25²

= –8.8 m/s²

As you can see using the second equation of motion is not that difficult. Should you enter into any difficulties just post the question into the comment section.

First equation of motion

The first equation of motion is mainly derived from the definition of acceleration.

You would remember that acceleration is the rate of change of velocity with time  and that the equation to calculate it is

acceleration = (change in velocity)/time taken

                    = (final velocity – initial velocity) /time taken

           a  = (v_f – v_i)/t

Rearranging the equation will give you

v_f = v_i + a*t

In some books the equation is given as such

v = u + at

If we used the equation above then the symbol v will be the final velocity and the symbol u will be the initial velocity of the object.

Now as we have seen in this post on acceleration if the final velocity is greater than the initial velocity then the acceleration is positive but if the initial velocity is greater than the initial velocity then the acceleration is negative.

Now let us have a look at two examples on the first equation of motion.

Example 1

A boy starts from rest accelerate to a velocity of 10 m/s in a time of 10 s. Determine the acceleration of the boy.

His final velocity v = 10 m/s

His initial velocity u = 0 m/s since he started from rest.

Time taken t = 10 s

Using the equation v = u +at and making  a the subject of formula will give you a = (v-u)/t

Substituting the different values of v, u and t in the equation will give you

a = (v-u )/t

  =(10 – 0 ) /10

  = 1 m/s²

Example 2

A train travelling at a speed of 100 m/s is decelerated to a speed of 50 m/s in a time of 15 s. Calculate the deceleration that the train is subjected.

final velocity v = 50 m/s

initial velocity u = 100 m/s

time taken t = 15 s

using the equation v = u +at and making a the subject of formula will give you the equation

a = (v-u) /t

Substituting in the equation above the different values of v, u and t

a = (v-u)/t

   = (50 –100 )/15

  = –50 /15

= - 3.33 m/s²

Since the acceleration is – 3.33 m/s² then

deceleration  = 3.33 m/s²

what is acceleration?

Acceleration is a term that is associated with a change with velocity of an object.

It is defined as the rate of change of velocity with time.

So it means that in order for an object to accelerate you need to have a change of velocity over a period of time.

 

There are three ways in which an object can accelerate that is its velocity can change over time.

1. Firstly as shown in fig 1 below the object’s velocity can increase with time.

Fig 1

2. Secondly as shown in fig 2 below the velocity of the object can decrease with time.

Fig 2

3. Lastly the velocity of the object can change due to a change in direction as shown below in fig 3. Remember the velocity is a vector quantity and as a result has a direction component and a magnitude component. Even if the magnitude remain constant a change in direction will result in a change in the velocity.

 

Fig 3

However we will only consider the first two cases today. The third one will be considered when we study circular motion.

If an object has a velocity v₁ is subjected to a force such that after a period of time t of being subjected to the force its velocity becomes v₂ then we can say that the object is experiencing an acceleration.

This acceleration can thus be calculated using the equation

From this equation we can thus see that when the velocity of an object increases then the acceleration is a positive number and when the velocity of the object decreases the acceleration is a negative number.

The acceleration is a vector quantity since it is a vector quantity divide by a scalar quantity.

What is power?

Power is the rate of doing work.

From the definition we can deduce the following equation

Power = Work done /Time taken

The unit of power is the Watt (symbol W) or the Joule/second (J/s)

Example 1

A boy pushes a box and as a result does 120 J of work in 10 s. What is the power developed by the boy?

Power = Work done / time taken

= 120 /10

= 12 W or J/s

There is another definition for power that is often used. It is

Power is the rate of dissipation of energy or the rate of change of energy conversion.

Power = Energy dissipated / Time taken

Example 2

A girl climbs a staircase gaining 500 J of gravitational potential energy in 10 s.

What is the power developed by the girl?

Power = Energy conversion / Time taken

= 500 / 10

= 50 W or J/s

Example 3

During the boiling of some water 4000 J of heat energy is dissipated in the kettle’s heater in a time of 8 s. What is the power of the heater?

Power = Energy dissipated/Time taken

= 4000/8

= 500 W or J/s

It is now time to do some questions. It will give these questions after a few of you have supplied your answers.

1. A trains of mass 50000 kg accelerated form rest and reaches a velocity of 50 ms^-1 in 60 s.

(a) Calculate the kinetic energy gained by the train.

(b) Calculate the power of the train engine.

2. A lamp is rated 80 W. If it is switched on for two hours, how much light energy is dissipated.