WEIGHT - Newton's Laws - SAT Physics Subject Test

SAT Physics Subject Test

Chapter 3 Newton”s Laws


For this test, remember that although they are used interchangeably in everyday life, mass and weight are not the same thing; there is a clear distinction between them in physics. The weight of an object is the gravitational force exerted on it by Earth (or by whatever planet on which it happens to be). Mass, in contrast, is an intrinsic property of an object that measures its inertia: An object”s mass does not change with location. If you put a baseball in a rocket and send it to the moon, its weight on the moon would be less than its weight on Earth (because the moon”s gravitational pull is weaker than Earth”s due to its much smaller mass), but its mass would be the same.

Since weight is a force, we can use Fnet = ma to compute it. What acceleration would gravitational force impose on an object? The gravitational acceleration, of course! Therefore, setting a = g, the equation Fnet = ma becomes

Fw = mg

This is the equation for the weight of an object of mass m (weight is often symbolized just by w, rather than Fw). Notice that mass and weight are proportional but not identical. Furthermore, mass is measured in kilograms, while weight is measured in newtons.

G or g?

Remember that G is a
universal constant equal to
6.67 × 10–11 Nxm2/kg2.
gdepends upon what
planet or moon an object
is on. Near the surface,
mg = where M and
R are the mass and radius
of the planet or moon.
Cancelling m, we see that g =

9. What is the mass of an object that weighs 500 N ?

Here”s How to Crack It

Since weight is m multiplied by g, mass is Fw (weight) divided by g. Therefore, m = Fw/g = (500 N)/(10 m/s2) = 50 kg.

10. A person weighs 200 pounds. Given that a pound is a unit of weight equal to 4.45 N, what is this person”s mass?

Here”s How to Crack It

This person”s weight in newtons is (200 lb)(4.45 N/lb) = 890 N, so his mass is m = Fw/g = (890 N)/(10 m/s2) = 89 kg.

11. A book whose mass is 2 kg rests on a table. Find the magnitude of the force exerted by the table on the book.

Here”s How to Crack It

The book experiences two forces: The downward pull of the earth”s gravity and the upward, supporting force exerted by the table. Since the book is at rest on the table, its acceleration is zero, so the net force on the book must be zero. Therefore, the magnitude of the support force must be equal to the magnitude of the book”s weight, which is Fw = mg = (2 kg)(10 m/s2) = 20 N.

12. A can of paint with a mass of 6 kg hangs from a rope. If the can is to be pulled up to a rooftop with an acceleration of 1 m/s2, what must be the tension in the rope?

Here”s How to Crack It

First draw a picture. Represent the object of interest (the can of paint) as a heavy dot, and draw the forces that act on the object as arrows connected to the dot. This is called a free-body (or force) diagram.

We have the tension force in the rope, FT (also symbolized merely by T), which is upward, and the weight, Fw, which is downward. Calling up the positive direction, the net force is FTFw. The second law, Fnet = ma, becomes FTFw = ma, so

FT = Fw + ma = mg + ma = m(g + a) = 6(10 + 1) = 66 N

13. A can of paint with a mass of 6 kg hangs from a rope. If the can is to be pulled up to a rooftop with a constant velocity of 1 m/s, what must be the tension in the rope?

Here”s How to Crack It

The phrase “constant velocity” automatically means a = 0 and, therefore, Fnet = 0. In the diagram above, FT would need to have the same magnitude as Fw in order for the can to be moving at a constant velocity. Thus, in this case, FT = Fw = mg = (6)(10) = 60 N.