### learning goals

At the end of the section, you can:

- Explain the difference between mass and weight.
- Explain why objects falling to Earth are never actually in free fall.
- Describe the concept of weightlessness.

Mass and weight are often used interchangeably in everyday life. For example, our medical records often show our weight in kilograms, but never in the correct unit, Newtons. In physics, however, there is an important difference. Weight is Earth's gravitational pull on an object. It depends on the distance from the center of the earth. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit or on the lunar surface.

### power units

The equation [Latex]{F}_{\text{net}}=ma [/latex] is used to define the net force in terms of mass, length and time. As explained above, the SI unit of force is the newton. Like [latex] {F}_{\text{net}}=ma, [/latex]

[Látex] 1\,\text{N}=1\,\text{kg}·{\text{m/s}}^{2}. [/Látex]

Although nearly everyone uses the newton as a unit of force, the most common unit of force in the United States is the pound (lb), where 1 N = 0.225 lb. Thus, a 225-lb person weighs 1000 N.

### weight and gravitational force

When an object is released, it accelerates towards the center of the Earth. Newton's second law states that a net force on an object is responsible for its acceleration. When drag is negligible, the net force on a falling object is the gravitational force, commonly referred to as its own force.**Weight**[latex] \overset{\to {w} [/latex], or its force due to gravity acting on an object of mass*Subway*. Weight can be called a vector because it has a direction;*sob*is, by definition, the direction of gravity, and therefore weight is a downward force. The magnitude of the weight is denoted as*c*. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration.*grama*. Using Galileo's result and Newton's second law, we can derive an equation for weight.

Imagine an object with mass*Subway*fall to the ground. It just experiences the downward gravitational force, i.e. the weight [latex] \overset{\to {w} [/latex]. Newton's second law states that the magnitude of the resultant external force on an object [latex] is {\overset{\to {F}}_{\text{net}}=m\overset{\to {a}. [/latex] We know that the gravitational acceleration of an object is [latex] \overset{\to {g}, [/latex] or [latex] \overset{\to {a}=\overset{ \a {g } [/latex]. Substituting them into Newton's second law, we get the following equations.

#### Weight

The gravitational force on a mass is its weight. We can write this in vector form, where [latex] \overset{\to }{w} [/latex] is the weight and*Subway*it's mass, how

[Látex] \overset{\to {w}=m\overset{\to {g}. [/Látex]

We can write in scalar form

[Látex] w=mg. [/Látex]

Since [Latex] g=9.80\,{\text{m/s}}^{2} [/latex] on Earth, the weight of a 1.00 kg object on Earth is 9.80 N :

[Latex] w=mg=(1.00\,\text{kg})({9.80\,\text{m/s}}^{2})=9.80\,\text{N} . [/Latex]

If the net external force on an object is its weight, we say it is within**free fall**, that is, the only force acting on the object is gravity. However, when objects on Earth fall, they are never truly in free fall, as there is always an upward aerodynamic force acting on the object.

Acceleration due to gravity*grama*varies slightly over the earth's surface, so an object's weight depends on its location and is not an intrinsic property of the object. Weight changes dramatically as we leave Earth's surface. For example, on the moon, the acceleration due to gravity is just [latex] {1.67\,\text{m/s}}^{2} [/latex]. A mass of 1.0 kg weighs 9.8 N on Earth and only about 1.7 N on the Moon.

The broadest definition of weight in this sense is that the weight of an object is the gravitational force exerted on it by the nearest large body, such as the Earth, Moon, or Sun. This is the most common and useful definition of weight in physics. However, it differs drastically from the definition of weight used by NASA and popular space and exploration-related media. When they talk about "weightlessness" and "microgravity" they are referring to the phenomenon we call "free fall" in physics. We use the above definition of weight, force [latex] \overset{\to {w} [/latex] due to the action of gravity on an object of mass*Subway*, and we carefully distinguish between free fall and true weightlessness.

Note that weight and mass are different physical quantities, although they are closely related. Mass is an intrinsic property of an object: it is a quantity of matter. The amount or amount of matter in an object is determined by the number of atoms and molecules of different types it contains. Since these numbers don't change, in Newtonian physics, mass doesn't change; therefore, its response to an applied force does not change. In contrast, weight is the gravitational force acting on an object, so it varies with gravity. For example, a person closer to the center of the Earth at a low altitude like New Orleans weighs slightly more than a person at the highest altitude in Denver, even though they might have the same mass.

It's tempting to equate mass with weight, since most of our examples take place on Earth, where an object's weight depends only slightly on its location. Also, it's difficult to count and identify all the atoms and molecules in an object, so mass is rarely determined this way. If we consider situations where [latex] \overset{\to {g} [/latex] is a constant on Earth, we see that the weight [latex] \overset{\to {w} [/latex] directly proportional to is the pasta*Subway*, since [latex] \overset{\to {w}=m\overset{\to {g}, [/latex] d. H. the more massive an object is, the more it weighs. Operationally, the masses of objects are determined by comparison with the standard kilogram, as discussed inUnits and Measures. But if we compare an object on Earth with one on the Moon, we can easily see a change in weight but not mass. For example, on Earth, a 5.0 kg object weighs 49 N; on the moon where*grama*is [latex]{1.67\,\text{m/s}}^{2} [/latex], the object weighs 8.4 N. However, the mass of the object on the moon is still 5.0 kg .

### Example

#### delete a field

A farmer lifts some medium weight stones from a field to plant. You are lifting a rock that weighs 40.0 lb (about 180 N). What force acts when the stone accelerates to a speed of [latex] 1.5\,{\text{m/s}}^{2}? [/Latex]

#### Strategy

They gave us the weight of the stone, which we used to determine the net force on the stone. However, we also need to know its mass to apply Newton's second law, so we need to apply the weight equation [latex] w=mg [/latex] to determine the mass.

#### Solution

There are no forces acting in the horizontal direction, so we can focus on the vertical forces as shown in the free-body diagram below. We denote the acceleration to the side; It's not technically part of the free-body diagram, but it helps us remember that the object is accelerating upward (so the net force is upward).

[latex] \begin{array}{ccc}\hfill w& =\hfill & mg\hfill \\ \hfill m& =\hfill & \frac{w}{g}=\frac{180\,\text{N} {9.8\,{\text{m/s}}^{2}}=18\,\text{kg}\hfill \\ \hfill \sum F& =\hfill & ma\hfill \\ \hfill F - w& =\fill & ma\fill \\ fill F-180\,\text{N}& =\fill & (18\,\text{kg})(1,5\,{\text{m/ s} } ^{2})\hfill \\\hfill F-180\,\text{N}&=\hfill&27\,\text{N}\hfill\\\hfill F&=\hfill&207\ ,\ text {N}=210\,\text{N with two significant digits}\hfill \end{array} [/latex]

#### Meaning

To apply Newton's second law as the main equation to solve a problem, sometimes we have to rely on other equations such as B. the weight equation or one of the kinematic equations to complete the solution.

### check your understanding

For(Example), find the acceleration when the force applied by the farmer is 230.0 N.

show solution

Can you avoid the debris field and land safely before you run out of fuel like Neil Armstrong did in 1969? OClassic video game versionAccurately simulates real Lunar Module motion with the correct mass, thrust, fuel burn rate, and lunar gravity. The real lunar module is difficult to control.

Use thisinteractive simulationMove the sun, earth, moon and space station around to see the effects on their orbits and gravitational forces. Visualize the sizes and distances between different celestial bodies and turn gravity off to see what would happen without it.

### problems

The weight of an astronaut plus spacesuit on the Moon is only 250 N. (a) How much does the astronaut's spacesuit weigh on Earth? (b) What is the mass of the moon? In the land?

show solution

Suppose the mass of a fully loaded module in which astronauts lift off from the moon is [latex] 1.00\,×\,{10}^{4} [/latex] kg. The thrust of their thrusters is [latex] 3.00\,×\,{10}^{4} [/latex] N. (a) Calculate the magnitude of the magnitude acceleration for a vertical launch from the moon. (b) Could it take off from Earth? If not, why not? If you can, calculate the magnitude of its acceleration.

A rocket sled accelerates at a speed of [latex] {49.0\,\text{m/s}}^{2} [/latex]. Your passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force that the seat exerts on your body. Compare this to your weight using a ratio. (b) Find the direction and magnitude of the total force the seat is exerting on your body.

show solution

Repeat the previous task for a situation where the rocket sled decelerates to a speed of [latex] {201\,\text{m/s}}^{2} [/latex]. In this problem, the forces are exerted by the seat and the seat belt.

A body with a mass of 2.00 kg is pushed upward by a vertical force of 25.0 N. What is its acceleration?

show solution

A car weighing 12,500 N starts from rest and accelerates to 83.0 km/h in 5.00 s. The frictional force is 1350 N. Find the applied force generated by the motor.

A body with a mass of 10.0 kg is in the Earth's gravitational field with [latex] g=9.80\,{\text{m/s}}^{2} [/latex]. What is your acceleration?

show solution

A firefighter has mass*Subway*; hears the fire alarm and slides down the pole with acceleration*A*(which is less than*grama*in magnitude). (a) Write an equation that gives the vertical force that must be exerted on the pole. (b) If your mass is 90.0 kg and you are accelerating at [latex] 5.00\,{\text{m/s}}^{2},[/latex], what is the applied force?

A baseball catcher performs a stunt for a television commercial. He picks up a baseball (mass 145 g) that is dropped from a height of 60.0 m above his glove. Your glove catches the ball in 0.0100 s. What force does your glove exert on the ball?

show solution

When the moon is directly overhead at sunset, the force from earth to moon, [latex] {F}_{\text{EM}} [/latex], is essentially [latex] 90\text{° } [ / latex ] to the force exerted by the sun on the moon, [latex] {F}_{\text{SM}} [/latex] as shown below. Data [latex] {F}_{\text{EM}}=1.98\,×\,{10}^{20}\,\text{N} [/latex] and [latex] {F}_ { \ text{SM}}=4.36\,×\,{10}^{20}\,\text{N}, [/latex] all other forces on the moon are negligible, and the mass of the moon is [ latex ] 7.35\,×\,{10}^{22}\,\text{kg}, [/latex] determines the magnitude of the lunar acceleration.