## Newtons Second Law of Motion: Net Force Causes Acceleration

Newton's Second Law of Motion: (Fnet = ma) All the net forces on one object cause it to accelerate if the net force is other than zero.

### Force and Acceleration Equation Variables

 Name Variable Unit Unit Abbreviation Force F Newton N Mass m kilogram kg Initial Velocity or Initial Speed vi meters per second m/s Final Velocity or Final Speed vf meters per second m/s Time t seconds s Acceleration a meters per second squared m/s2 Displacement or Distance X meters m

### Force Diagram Versus Free Body Diagram

We use a force diagram to show all forces with direction acting on an object.  These forces are drawn from a center point if a picture of the object is drawn.  Free body diagrams also have the forces involved drawn but a dot represents the object instead.  Free body force diagram means no object is drawn in a force diagram.

In a later section you will see that weight equals mass times acceleration (FW = mg).  If given mass (in kilograms (kg)) you must convert to weight (in Newtons (N)) using the 9.8 m/s2 acceleration due to gravity on earth.  On a horizontal surface normal force, the force the ground pushes back on an object is equal to weight and pushing up.  Weight (FW) is also the minimum force a person would have to apply to lift an object up once moving.

Newton's Second Law is simply stated as an equation:

## Fnet = ma

You need to take into account all force vectors acting on an object (as seen to the right) and use the net force to determine the acceleration.

Note: If there is only one force mentioned in a problem then this would be your net force and the equation above is simply stated F = ma.

Acceleration is directly related to force.  A greater force gives you more acceleration.  When force goes up acceleration goes up.

Acceleration is inversely related to mass.  Look at the animation to the right.  When mass goes up acceleration goes down.

### Problems Involving Newtons Second Law: Force and the One Dimensional Motion Equation

Problems involving the net force equation (Fnet = ma) often tie into one dimensional motion problems.  Notice in the picture to the right that you may have force and mass to solve for acceleration.  Later you may use this to find the final velocity or another unknown seen in the left side of the picture.

You may also have to start with the equations on the left.  Later you may use those to find a mass or net force causing that acceleration.

Making a givens list will help you determine how to proceed in a problem.

### Example Problems

1. What is the net force and how much would a 2 kg object accelerate when you have a 12 N force right and 6 N force left?

Turn directions into signs (I'll call right positive for these when one dimensional: on the same axis)

+ 12 N and - 6 N

(12) + (-6) = +6 N

Turn the sign back into a direction

6 N right

Use the net force to calculate acceleration equation

Fnet­  = ma

a = Fnet­  / m

a = 6 / 2 = 3 m/s2 right

2. What is the net force and how much would a 2 kg object accelerate when you have a 12 N and 6 N force both left?

Turn directions into signs

- 12 N and - 6 N

(-12) + (-6) = +18 N

Turn the sign back into a direction

18 N left

Use the net force to calculate acceleration equation

Fnet­  = ma

a = Fnet­  / m

a = 18 / 2 = 9 m/s2 left

Acceleration is always in the same direction of net force

3. What is the net force and how much would a 2 kg object accelerate when you have a 12 N force up and another 6 N left?

Since the vectors are in right angles so you have to follow these steps:

• First: Set up the vectors head to tail, draw the resultant, and place the angle closest to the origin.  The beginning of the first arrow is the origin.
• Second: Use Pythagorean Theorem to find the magnitude and inverse tangent to find the angle and then describe it.

The resultant force would be 13.4 N 26.6° above the left horizon.  (There are many ways to describe the angle when not really defined as long as you have the baseline and an angle in a direction from it.)

Find the Acceleration which is in the same direction as net force

Fnet­ = ma

a = Fnet­ / m

a = 13.4/ 2

a = 6.7 m/s2 above the left horizon

### Example Problems

4. A force of 1120 N is applied to a go-cart with a mass of 510 kg.  How fast will the go-cart be traveling after 4 seconds from rest?

Solution:

Givens List:

F = 1120 N     m = 510 kg     vi = 0     vf = ?     t = 4s

You have force and mass and start by solving for acceleration so you can use it in a one dimensional motion acceleration equation.  Since there is only one force mentioned it is your Fnet.

F­ = ma

a = F­ / m

a = 1120 / 510 = 2.20 m/s2

Now that you have acceleration pick from your acceleration equations using the following givens and unknown?

Givens List:

a = 2.20 m/s2        vi = 0        vf = ?        t = 4s

Equation and Solution:

vf = vi + at

vf = 0 + (2.2)(4) = 8.8 m/s

The final velocity will be 8.8 m/s forward.

(If you need more help picking your equation go back to our 1D accelerated motion lesson here)

### 5. How much force is required to push a 6.5 kg shopping cart 10 meters in 3 seconds starting from rest?

Solution:

Start by using x =10 m, t = 3s, vi = 0, to solve for a = ?

Pick the right equation and solve

x = vit + ½ at2

Rearranges to

Now solve for the force required to accelerate a 6.5 kg shopping cart by 2.22 m/s2

F = ma

F = (6.5)(2.22) = 14.4 N forward

### Example Problem (Force Applied at an Angle)

When force is applied at an angle, only the force component in the direction of motion will directly accelerate the object.  The horizontal component, the adjacent side of the triangle to the right, will accelerate the 5 kg mass.

6. How much would a 5 kg box accelerate when a 75 N force is applied 25° above the horizon on the right side?

You would start by finding the adjacent side:

Adj = (cos25)(75) = 68 N

Then figure out the acceleration using Fnet=ma:

F = ma

a = F/m

a = 68/5 = 13.6 m/s2 right without friction present

### Example Problem

7. What is the normal force created by the ground on a 5 kg box when a 75 N force is applied 25° above the horizon? (see diagram for #6)

Normal force will be important in a later section when we need to solve for the force of friction (Ff = µFN)

Normal force is the force the ground pushes up against the weight directed down created by gravitational force.  When pulling up at an angle, the vertical component of force will subtract from weight and normal force up.

Begin by calculating weight:

FW = mg

FW = (5)(9.8) = 49 Newton’s

Normal Force (FN) would have been 49 N without the pulling up component of this picture.

Now determine the vertical or upward component of the force:

Opp = sin (25)(75) = 31.7 N

Now determine the net force (overall force):

The normal force in this example must make the net force 0 in the Y axis since the box is not moving up or down.

0 Net force = 49 N down + 31.7 N up + F­N up

0 = -49 N + 31.7 N + F­N

FN = 17.3 N

Newton's Second Law Quiz

1 / 10

Which of Newton's Laws is most related to a larger object resisting a change of motion more than a less massive one?

2 / 10

Which of Newton's Laws is most related to all the forces on an object causing the object to accelerate?

3 / 10

Force and acceleration are ______________ related

4 / 10

Acceleration and mass are ______________ related

5 / 10

What is the mass of an object that accelerates at 4 m/s when 20 N of net force is applied?

6 / 10

What is the net force in the picture above

7 / 10

What is the acceleration on a 5 kg object that has a force of 20 N applied to the right and 5 newtons applied to the left?

8 / 10

What is the net force on a 5 kg object that has a force of 20 N applied to the right and 5 newtons applied to the left?

9 / 10

Net force causes _____________________.  Pick the best answer

10 / 10

Which direction does an object accelerate?