Download Newton`s Second Law File

Newton’s Second Law
Add Important
Notes/Cues Here
Page: 1
Unit: Forces
Newton’s Second Law
Unit: Forces
NGSS Standards: HS-PS2-1
MA Curriculum Frameworks (2006): 1.4
Skills:
 Solve problems relating to Newton’s Second Law (F = ma)
 Solve problems that combine kinematics (motion) and forces
Language Objectives:
 Identify and correctly use the quantities involved in a Newton’s Second Law
problem.
 Identify and correctly use the quantities involved in a problem that
combines kinematics and forces.
Notes:
Newton’s Second Law: Forces cause acceleration (a change in velocity). “A net

force, F , acting on an object causes the object to accelerate in the direction
of the net force.”
If there is a net force, the object accelerates (its velocity changes). If there is
no net force, the object’s velocity remains the same.
If an object accelerates (its velocity changes), there was a net force on it. If
an object’s velocity remains the same, there was no net force on it.
Remember that forces are vectors. No net force can mean no forces at all, or it
can mean that there were equal forces in opposite directions and their effects
cancel.
Use this space for summary and/or additional notes.
Copyright © 2010–2017 Mr. Bigler.
Physics
This document is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported
License. This license gives you permission to copy, share and/or adapt these works, with
appropriate attribution, under an identical, similar, or compatible license. See
http://creativecommons.org/licenses/by-sa/3.0/ for more information.
Mr. Bigler
Newton’s Second Law
Add Important
Notes/Cues Here
In equation form:
Page: 2
Unit: Forces

 Fnet
a
m
This form is preferred, because acceleration is what results from a force applied
to a mass. (I.e., force and mass are the independent variables and acceleration is
the dependent variable. Forces cause acceleration, not the other way around.)
However, the equation is more commonly written:


Fnet  ma
Sample Problems
Most of the physics problems involving forces require the application of


Newton’s Second Law, Fnet  ma .
Q: A net force of +50 N is applied to a cart that has a mass of 35 kg. How fast
does the cart accelerate?
A: Applying Newton’s Second Law:


F  ma

 50  35 a

a  1.43 sm2
Q: What is the weight of (i.e., the force of gravity acting on) a 10 kg block?


A: F  mg

F  (10. )(9.8)  98 N


(Remember that we use the variable g instead of a when the acceleration is
caused by gravity.)
Use this space for summary and/or additional notes.
Physics
Mr. Bigler
Newton’s Second Law
Add Important
Notes/Cues Here
Page: 3
Unit: Forces
Free Body Diagrams and Newton’s Second Law
Free-body diagrams are often used in combination with Newton’s second law


( Fnet  ma ); the free-body diagram enables you to calculate the net force, from
which you can calculate mass or acceleration.
Sample Problem:
Q: Two children are fighting over a toy.
Jamie pulls to the left with a force of 40 N, and Edward pulls to the right with
a force of 60 N. If the toy has a mass of 0.6 kg, what is the resulting
acceleration of the toy?
A: Let us decide that the positive direction is to the right. (This is convenient
because the force to the right is larger, which means the net force will come
out to a positive number.)
The force diagram looks like this:


Fnet  ma

 20  (0.6) a
 20
a
 33.3 sm2 (to the right)
0.6
Use this space for summary and/or additional notes.
Physics
Mr. Bigler
Add Important
Notes/Cues Here
Newton’s Second Law
Page: 4
Unit: Forces
Q: A 5.0 kg block is resting on a horizontal, flat surface. How much force is
needed to overcome a force of 2.0 N of friction and accelerate the block from
rest to a velocity of 6.0 ms over a 1.5-second interval?
A: This is a combination of a Newton’s second law problem, and a motion
problem. We will need a free-body diagram to be able to visualize what’s
going on.
The free-body diagram for the block looks like this:
1. The net force is given by:
Fnet  Fa  Ff  Fa  2
Fa  Fnet  2
This means we need to find the net force, and then add 2 N to get the
applied force.
2. To find the net force, we need the equation:
Fnet  ma
3. We know that m = 5.0 kg, but we don’t know a. We need to find a in
order to calculate Fnet . For this, we will turn to the motion problem.
4. The problem tells us that v o  0 , v  6.0 ms , and t  1.5 s . Looking at the
motion equations, we see that we have all of the variables except for a in
the equation:
v  v o  at
Use this space for summary and/or additional notes.
Physics
Mr. Bigler
Add Important
Notes/Cues Here
Newton’s Second Law
Page: 5
Unit: Forces
3. Our strategy is to solve this equation for a, then substitute into Fnet  ma
to find Fnet , then use the relationship we found from the free-body
diagram to find Fa .
The acceleration is:
v  v o  at
6.0  0  a (1.5)
6.0  1.5 a
a  4.0 sm2
Substituting back into Fnet  ma gives:
Fnet  ma
Fnet  (5.0)(4.0)
Fnet  20. N
Finally:
Fa  Fnet  2
Fa  20  2
Fa  22 N
Problems like this are straightforward to solve, but they are challenging because
you need to keep chasing the quantities that you don’t know until you have
enough information to calculate them. However, you need to keep track of each
step, because once you have found the last equation you need, you have to
follow the steps in reverse order to get back to the answer.
Use this space for summary and/or additional notes.
Physics
Mr. Bigler