# Newton's Second Law By: Matt Gram © & Thai Thov ® , December1997

## 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)

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

## 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

## 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

## 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

## 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

## 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

## 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

## 8.

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

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

## 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

## 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

## 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

## 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