Newton's Second Law


By: Matt Gram © & Thai Thov ® , December1997

Table of Contents


Here are the quantities that are used in this lesson:

These quantities are also defined and explained on other pages



What is Weight?



Weight can be defined by the gravitational force on an object, with a known mass. This means that in order to obtain the weight of an object you take the mass of an object, and multiply that by the acceleration of gravity. The mass of an object is the same no matter where it exists, while the weight of an object changes depending on what the gravitational field strength where the object exists. Weight is measured in Newtons (N).



Converting to Kilograms


The prefix of kilogram is kilo-. Which means 1000. Therefore a kilogram is one thousand grams.

Example

Convert 5 grams to kilograms

5 grams/1000

m = .005 grams

Example 2

Convert .008 Kg to grams

(.008) * (1000) = m

m = 8 grams


Go back to: Table of Contents


Formulas

Here are the formulas that you need to memorize and use for this lesson:

Go back to: Table of Contents

General Problem Solving Strategy:

  1. Read the problem.
  2. Go through the problem and figure out what is given or implied.
  3. Make a list, and identify the quantities you know.
  4. Convert to grams or kilograms if necessary.
  5. Use one of the two formulas to calculate anything you don't know.
  6. Put what you have just solved for on you list of know variables.
  7. If the problem is not finished, use the appropriate formula to solve for the needed variable.
  8. If you are not able to solve a problem, reread the problem to see if you used the correct values.


Go back to: Table of Contents

Example problem 1


What is the weight of a 2.7 gram penny on Earth and on the moon? (Gravitational force on the moon is 1/6 that of the Earth.)


Here is what you know:

    


  1. m = 2.7 grams
  2. g = 9.8 m/s/s

Now you must convert 2.7 grams to kilograms.

2.7 grams/1000 = m

m = .0027 Kg


Now you have enough information to use the second formula. (W = mg)

w = m * g

w = .0027 grams * 9.8 m/s/s

w = .02646 N

This is the weight of a penny in Newton's on Earth.


Now you must find the weight of a penny on the moon.


Here is what you start with:

    


  1. m = .0027 Kg
  2. g = 1/6 * 9.8 m/s/s

You have enough information to solve the problem with the second formula.

w = mg

w = . 0027 Kg * 1/6(9.8 m/s/s)

w = .0044 N


Problem solved.


Go back to: Table of Contents

Example problem 2


A 3000 Kg plane has an acceleration of 12 m/s/s on take off, what net force is excerted on the plane?


Here is what you know:

    


  1. m = 3000 Kg
  2. a = 12 m/s/s

From the information given, you can solve the problem by using the first equation. (F = ma)


Here is the work that should be done.

F = m * a

F = 3000 Kg * 12 m/s/s

F = 36,000 N

Problem solved.
Go back to: Table of Contents

Sample Problems



The answers to each problem follow it in parenthesis.  They also link to a solution to the


problem.  Try the problem, check your answer, and go to the solution if you do not understand.

1.

What is the acceleration of a 4.5 kg mass when there is a net force of 18 N on it?

4m/s/s

2.

A space worker exerts a net force of 256 N on an object and it accelerates at 8 m/s/s. What is the mass of the object?

32 Kg

3.

Sally accelerates a 250 Kg cart at 3 m/s/s. What must be the net force?

750 N

4.

What must be the mass of a model rocket in grams to develop an acceleration of 520 m/s/s when subjected to a net force of 11.2 N

21.5 grams

5.

What force can stop a 1400 Kg truck traveling at 30 m/s in 4 seconds?

10500 N

Go back to: Table of Contents.

6.

What force can stop the truck in .2 seconds?

210,000 N

7.

A car can go from o to 27 m/s in 4.5 seconds. If a net force of 6600 N acted on the car what is its mass?

1100 Kg

8.

How much does a 60 Kg person weigh?

588 N

9.

How much does a 392 g object weigh?

3.84 N

10.

What is the mass of an object that weighs 720 N on the surface of the earth?

73 Kg

11.

What is the strength of the gravitational force around a planet where a 45 Kg object weighs 18 N?

.4 N/Kg

12.

What is the acceleration of a 19.6 N object (on earth) that experiences a net force of 8 N?

4 m/s/s

Go to: Table of Contents

Solutions to sample problems

1.

 What is the acceleration of a 4.5 kg mass when there is a net force of 18 N on it? (4 m/s/s)

This is the obvious information you start with:

    


  1. m = 4.5 Kg
  2. F = 18 N

You have enough information to use the first formula. Using it properly you will arrive at the answer.


Formulas

F = ma

18 N = 4.5 Kg * a

a = 18 N/4.5 Kg

a = 4 m/s/s

Go back to: Table of Contents Problem 1

2.

 A space worker exerts a net force of 256 N on an object and it accelerates at 8 m/s/s. What is the mass of the object?
(32 Kg)

This is the given information: 

    


  1. F = 256 N
  2. a = 8 m/s/s

By looking at the given information, you know force, and an acceleration. Therefore you have enough information to use the first formula.


F = ma

256 N = m * 8 m/s/s

m = 256 N/8 m/s/s

m = 32 Kg


Go back to: Table of Contents Problem 2

3.

 Sally accelerates a 250 Kg cart at 3 m/s/s. What must be the net force? (750 N)

This is the given information:

    


  1. m = 250 Kg
  2. a = 3 m/s/s

By looking at the given information, you know a mass and an acceleration. You are given enough information to use the first formula.


F = ma

F = 250 Kg * 3 m/s/s

F = 750 N


Go back to: Table of Contents Problem 3

4.

 What must be the mass of a model rocket in grams to develop an acceleration of 520 m/s/s when subjected to a net force of 11.2 N. (21.5 Grams)

This is the given information:

    


  1. F = 11.2 N
  2. a = 520 m/s/s

The given information gives you a force and an acceleration. You have enough information to use the first fomula.


F = ma

11.2 N = m * 520 m/s/s

m = 11.2 N/520 m/s/s

m = .0215 Kg


The question asked for mass in grams. The mass right now is in kilograms and must be converted.

Since there are 1000 grams in a kilogram, you must multiply your answer by 1000 to convert to grams.


.0215 Kg * 1000 grams = 21.5 grams

m = 21.5 grams


Go back to: Table of Contents Problem 4

5.

 What force can stop a 1400 Kg truck traveling at 30 m/s in 4 seconds? (10,500 N)

This is the given information:

    


  1. m = 1400 Kg
  2. V = 30 m/s
  3. t = 4 sec

The problem gives you a mass, a velocity, and a time. That is not enough information to use a formula.

You should recognize from previous lessons change in that velocity divided by time, equals acceleration.


a = V/t

a = 30 m/s / 4 sec

a = 7.5 m/s/s

These are the variables that you now know.

    


  1. m = 1400 Kg
  2. V = 30 m/s
  3. t = 4 sec
  4. a = 7.5 m/s/s

You now have enough information to use the first formula.

F = ma

F = 1400 Kg * 7.5 m/s/s

F = 10,500 N


Go back to: Table of Contents Problem 5

6.

 What force can stop the truck in .2 seconds? (210,000 N)

This problem is a continuation of problem # 5. All the information except for time, and acceleration, is equal to problem 5.

This is what you know: 
    


  1. m = 1400 Kg
  2. V = 30 m/s
  3. t = .2 sec

The problem gives you a mass, a velocity, and a time. That is not enough information to use a fomula.

You should recognize from previous lessons that velocity divided by time, equals acceleration.


a = V/t

a = 30 m/s / .2 sec

a = 150 m/s/s

These are the variables that you now know.

    


  1. m = 1400 Kg
  2. V = 30 m/s
  3. t = 4 sec
  4. a = 150 m/s/s

You now have enough information to use the first formula.

F = ma

F = 1400 Kg * 150 m/s/s

F = 210,000 N


Go back to: Table of Contents Problem 6

7.

 A car can go from o to 27 m/s in 4.5 seconds. If a net force of 6600 N acted on the car what is its mass? (1100 Kg)

This is the given information:

    


  1. F = 6600 N
  2. V = 27 m/s
  3. t = 4.5 sec

The problem gives you a force, a velocity, and a time. That is not enough information to use a formula.

You should recognize from previous lessons that velocity divided by time, equals acceleration.


a = V/t

a = 27 m/s / 4.5 sec

a = 6 m/s/s

These are the variables that you now know.

    


  1. F = 6600 N
  2. V = 27 m/s
  3. t = 4.5 sec
  4. a = 6 m/s/s

You now have enough information to use the first formula.

F = ma

6600 N = m * 6 m/s/s

m = 6600 N/6 m/s/s

m = 1100 Kg


Go back to: Table of Contents Problem 7

8.

 How much does a 60 Kg person weigh? (588 N)

This is what you start with:

    


  1. m = 60 Kg
  2. g = 9.8 m/s/s

Gravity is equal to 9.8 m/s/s because the question does not specify where the person is located. So assume the person is on Earth.

You have enough information to use the second formula.

Formulas

w = mg

w = 60 Kg * 9.8 m/s/s

w = 588 N


Go back to: Table of Contents Problem 8

9.

 How much does a 392 g object weigh? (3.84 N)

This is the given information:

    


  1. m = 392 g
  2. g = 9.8 m/s/s

The mass is given in grams. In order to use the second equation it should be in kilograms, so you must convert.

m = 392 g/1000

m = .392 Kg

This is what you now know:

    


  1. m = .392 Kg
  2. g = 9.8 m/s/s

You now have enough information to use the second equation.

w = mg

w = .392 Kg * 9.8 m/s/s

w = 3.84 N


Go back to: Table of Contents Problem 9

10.

 What is the mass of an object that weighs 720 N on the surface of the earth? (73 Kg)

This is the given information:

    


  1. W = 720 N
  2. g = 9.8 m/s/s

    3. Enough information is given to use the second formula.

    w = mg

    720 N = m * 9.8 m/s/s

    m = 720 N/9.8 m/s/s

    m = 73 Kg


    Go back to: Table of Contents Problem 10

    11.

    What is the strength of the gravitational force around a planet on a 45 Kg object weighs 18 N? (.4 N/Kg)

    This is the information given:

      


    1. m = 45 Kg
    2. w = 18 N

    You have enough information to use the second formula.

    w = mg

    18 N = 45 Kg * g

    g = 18N/45 Kg

    g = .4 n/Kg


    Go back to: Table of Contents Problem 11

    12.

     What is the acceleration of a 19.6 N object (on earth) that experiences a net force of 8 N? (4M/S/S)

    This is the information you know.

      


    1. w = 19.6 N
    2. F = 8 N
    3. g = 9.8 m/s/s

    Since the object weighs 19.6 N, and it is on earth, you can use the second formula to find it's mass.

    w = mg

    19.6 N = m * 9.8 m/s/s

    m = 19.6 N/9.8 m/s/s

    m = 2 Kg

    This is the information you have now:

      


    1. w = 19.6 N
    2. F = 8 N
    3. g = 9.8 m/s/s
    4. m = 2 Kg

    Now you have enough information to use the first formula and finish the problem.

    F = ma

    8 N = 2 Kg * a

    a = 8 N/2 Kg

    a = 4 m/s/s


    Go back to: Table of Contents Problem 12