Newton's Second Law: formula, examples and exercises

Newton's Second Law establishes that the acceleration acquired by a body is directly proportional to the resultant of the forces acting on it.

As acceleration represents the variation in velocity per unit of time, the 2nd Law indicates that forces are the agents that produce variations in velocity in a body.

Also called the fundamental principle of Dynamics, it was conceived by Isaac Newton and forms, together with two other laws (1st Law and Action and Reaction), the foundations of Classical Mechanics.

Formula

We mathematically represent the Second Law as:

stack F with R subscript with right arrow above equal to m space. a space with right arrow superscript

Where,

stack F with R subscript with right arrow above two points space fo r ç a space r e s u l tan t e. space A space u n i d e space n the space s i s t m a space i n t e r n a c i o n a l space is space the space n and w t o n space left parenthesis N right parenthesis.
m colon space m a s s a. space A space u n i d e space n the space s i s t m a space i n t e r n a c i o n a l space is space the space q u i log r a m a space left parenthesis k g right parenthesis.
a with right arrow superscript colon space a c e l e r ation. space A space un i d e space n space S I narrow space is space the space m e tr space for space s e g u n d the space a the space q u a d r a d the space left parenthesis m divided by s squared right parenthesis

Force and acceleration are vector quantities, so they are represented with an arrow over the letters that indicate them.

As vector quantities, they need, in order to be fully defined, a numerical value, a unit of measure, a direction and a direction. The direction and direction of acceleration will be the same as the net force.

In the 2nd Law, the object's mass (m) is the proportionality constant of the equation and is the measure of the inertia of a body.

In this way, if we apply the same force on two bodies with different masses, the one with the greatest mass will suffer a lower acceleration. Hence, we conclude that the one with greater mass resists more to variations in velocity, therefore, it has greater inertia.

Newton's Second Law
Force equals mass times acceleration

Example:

A body with a mass equal to 15 kg moves with a modulus acceleration equal to 3 m/s2. What is the magnitude of the net force acting on the body?

The force module will be found applying the 2nd law, so we have:

FR = 15. 3 = 45 N

Newton's Three Laws

the physicist and mathematician Isaac Newton (1643-1727) formulated the basic laws of mechanics, where he describes the movements and their causes. The three laws were published in 1687, in the work "Mathematical Principles of Natural Philosophy".

Newton's First Law

Newton was based on the ideas of Galileo on inertia to formulate the 1st Law, therefore, it is also called the Law of Inertia and can be stated:

In the absence of forces, a body at rest remains at rest and a body in motion moves in a straight line with constant velocity.

In short, the Newton's First Law indicates that an object cannot initiate motion, stop or change direction by itself. It takes the action of a force to bring about changes in its state of rest or movement.

Newton's Third Law

THE Newton's Third Law it is the Law of "Action and Reaction". This means that, for every action, there is a reaction of the same intensity, same direction and in the opposite direction. The principle of action and reaction analyzes the interactions that take place between two bodies.

When a body suffers the action of a force, another will receive its reaction. As the action-reaction pair occurs in different bodies, the forces do not balance.

Learn more at:

  • Newton's Three Laws
  • Gravity
  • What is Inertia in Physics?
  • Physics Formulas
  • Quantity of Movement
  • inclined plane

Solved Exercises

1) UFRJ-2006

A block of mass m is lowered and raised using an ideal wire. Initially, the block is lowered with constant vertical acceleration, downward, of module a (by hypothesis, less than the module g of the acceleration of gravity), as shown in figure 1. Then, the block is lifted with constant vertical acceleration, upwards, also of module a, as shown in figure 2. Let T be the yarn tension on the way down and T’ the yarn tension on the way up.

UFRJ Exercises

Determine the ratio T’/T as a function of a and g.

In the first situation, as the block is descending, the weight is greater than the traction. So we have that the net force will be: FR=P - T
In the second situation, when going up T' will be greater than the weight, so: FR=T' - P
Applying Newton's 2nd law, and remembering that P = m.g, we have:
left parenthesis 1 right parenthesis P space minus T space equal to m space. a space double right arrow T equal to m. g space minus m space. The
left parenthesis 2 right parenthesis space T apostrophe minus P space equal to m. double right arrow T apostrophe equals m. the most m. g
Dividing (2) by (1), we find the requested reason:
numerator T ´ over denominator T end of fraction equals numerator g space plus a over denominator g minus end of fraction

2) Mackenzie-2005

A 4.0kg body is being lifted by means of a wire that supports maximum traction of 50N. Adopting g = 10m/s2, the greatest vertical acceleration that can be applied to the body, pulling it by this wire, is:

a) 2.5m/s2
b) 2.0m/s2
c) 1.5m/s2
d) 1.0m/s2
e) 0.5m/s2

T - P = m. a (the body is being lifted, so T>P)
As the maximum traction is 50 N and P = m. g = 4. 10 = 40 N, the greatest acceleration will be:
50 minus 40 equals 4. the double right arrow a equals 10 over 4 equals 2 comma 5 m space divided by s squared

Alternative to: 2.5 m/s2

3) PUC/MG-2007

In the figure, block A has a mass mTHE = 80 kg and block B, a mass mB = 20 kg. Frictions and inertia of the wire and pulley are still negligible and g = 10m/s is considered2 .

blocks

Regarding the acceleration of block B, it can be said that it will be:

a) 10 m/s2 down.
b) 4.0 m/s2 up.
c) 4.0 m/s2 down.
d) 2.0 m/s2 down.

B's weight is the force responsible for moving the blocks down. Considering the blocks as a single system and applying Newton's 2nd Law we have:
PB = (mTHE + mB). The
a equals numerator 20.10 over denominator 80 plus 20 end of fraction equals 200 over 100 equals 2 m space divided by s squared

Alternative d: 2.0 m/s2 down

4) Fatec-2006

Two blocks A and B of masses 10 kg and 20 kg, respectively, joined by a thread of negligible mass, are at rest on a horizontal plane without friction. A force, also horizontal, of intensity F = 60N is applied to block B, as shown in the figure.

blocks

The modulus of the traction force in the wire that joins the two blocks, in newtons, is valid

a) 60
b) 50
c) 40
d) 30
e) 20

Considering the two blocks as a single system, we have: F = (mTHE + mB). a, substituting the values ​​we find the acceleration value:

a equals numerator 60 over denominator 10 plus 20 end of fraction equals 60 over 30 equals 2 m space divided by s squared

Knowing the value of the acceleration, we can calculate the value of tension in the wire, let's use block A for this:

T=mTHE . The
T = 10. 2 = 20 N

Alternative e: 20 N

5) ITA-1996

Shopping in a supermarket, a student uses two carts. It pushes the first one, of mass m, with a horizontal force F, which, in turn, pushes another one of mass M onto a flat, horizontal floor. If the friction between the carts and the floor can be neglected, it can be said that the force that is applied on the second cart is:

a) F
b) MF / (m + M)
c) F (m + M) / M
d) F / 2
e) another different expression

Considering the two carts as a single system, we have:

F equals left parenthesis m plus M right parenthesis space. space right double arrow a equals numerator F over denominator left parenthesis m plus M right parenthesis end of fraction

To calculate the force acting on the second cart, let's use Newton's 2nd Law for the 2nd cart equation again:

f equals M space. space the double right arrow f equals M. numerator F over denominator left parenthesis m plus M right parenthesis end of fraction

Alternative b: MF/(m+M)

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