Home Page Textbooks Physics: Principles and Problems Page 101: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 101: Section Review

Exercise 21

Step 1

1 of 3

In this task, we will compare the gravitational forces on Earth and the Moon. We will do this by calculating the ratio of these two forces.
Known:

$$
begin{align*}
m&=10 mathrm{kg} \
g_{Earth}&=9.8 mathrm{m/s^2} \
g_{Moon}&=1.6 mathrm{m/s^2}
end{align*}
$$

Unknown:

$$
begin{align*}
F_{g(Earth)}/F_{g(Moon)}&=?
end{align*}
$$

Step 2

2 of 3

To calculate the ratio, let us recall the expression for calculating the gravitational force.

$$
begin{align*}
F_g&=mg \
\
F_{g(Earth)}/F_{g(Moon)}&=dfrac{mg_{Earth}}{mg_{Moon}} \
&=dfrac{g_{Earth}}{g_{Moon}} \
&=dfrac{9.8 mathrm{m/s^2}}{1.6 mathrm{m/s^2}} \
&=boxed{6.125}
end{align*}
$$

From the force ratio we can see that the force of gravity on the Moon is approximately 6 times smaller than the force of gravity on Earth.

Result

3 of 3

Force of gravity is $6.125$ less than force of gravity on Earth.

Exercise 22

Step 1

1 of 2

Let the apparent weight be the force shown by the scale. The apparent and actual weight are the same when the elevator is moving at a constant speed, the free body diagram for this case is shown under case 1. When the elevator slows down when reaching the top and accelerates when returning down, the apparent weight is less than the actual. The free body diagram for this situation is shown under case 2. Finally, in the case where the elevator accelerates upwards and slows downwards, i.e. at the beginning and end of the movement, the apparent weight is greater than the actual one. The free body diagram for the last situation is under case 3. Exercise scan

Result

2 of 2

The apparent weight is greater than the actual at the very beginning and end of the movement, less than the actual when coming and going from the top, and equal to, during travel at the constant speed.

Exercise 23

Step 1

1 of 4

In this problem we want to calculate the acceleration for a known amount of force and mass.
Known:

$$
begin{align*}
F&=9 mathrm{N} \
m&=65 mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
a&=?
end{align*}
$$

Step 2

2 of 4

We calculate acceleration from Newton’s second law:

$$
begin{align*}
F&=mcdot a \
a&=dfrac{F}{m}
end{align*}
$$

Step 3

3 of 4

$$
begin{align*}
a&=dfrac{F}{m} \
&=dfrac{9 mathrm{N}}{65 mathrm{kg}} \
&=boxed{0.1385 mathrm{m/s^2}}
end{align*}
$$

Result

4 of 4

$$
a=0.1385 mathrm{m/s^2}
$$

Exercise 24

Step 1

1 of 5

In this task we observe what happens when we hold a weight on a spring while riding an elevator. Consider what we can conclude from the known data.

Known:

$$
begin{align*}
m&=1 mathrm{kg} \
F_s&=9.3 mathrm{N} \
g&=9.8 mathrm{s^2}
end{align*}
$$

Step 2

2 of 5

When we are in an elevator, a force of gravity acts on the weight, which pulls it down. Let’s calculate what the weight of the weight is.

$$
begin{align*}
F_g&=mcdot g \
&=1 mathrm{kg} cdot 9.8 mathrm{m/s^2} \
&=9.8 mathrm{N}
end{align*}
$$

Step 3

3 of 5

We can see that the weight is greater than the force shown by the scales. It is possible that the elevator accelerates and that due to the inertial force the scale shows a lower value. Acceleration can be calculated by pluging in the difference in weight and force shown by the scale in Newton’s second law.

$$
begin{align*}
F&=mcdot a \
F_s-F_g&=mcdot a \
\
a&=dfrac{F_s-F_g}{m} \
&=dfrac{9.3 mathrm{N}-9.8 mathrm{N}}{1 mathrm{kg}} \
&=boxed{-0.5 mathrm{m/s^2}}
end{align*}
$$

Step 4

4 of 5

If we determine that the positive upward direction the elevator moves with acceleration $a=-0.5 mathrm{m/s^2}$. This means that it either accelerates downwards or slows downwards.

Result

5 of 5

If the upward direction is positive, the elevator moves with acceleration $a = -0.5 mathrm{m/s^ 2}$.

Exercise 25

Step 1

1 of 4

In this task we want to determine the mass of a toy with known amounts of force exerted on the toy and the acceleration of the toy.

Known:

$$
begin{align*}
F_{Marcos}&=22 mathrm{N} \
F_{cat}&=-19.5 mathrm{N} \
a&=6.25 mathrm{m/s^2}
end{align*}
$$

Unknown:

$$
begin{align*}
m&=?
end{align*}
$$

Step 2

2 of 4

We choose toward Marcos as positive direction. In order to calculate the mass of a toy, we must first calculate the net force on the toy.

$$
begin{align*}
F_{net}&=F_{Marcus}+F_{cat} \
&=22 mathrm{N}-19.5 mathrm{N} \
&=2.5 mathrm{N}
end{align*}
$$

Step 3

3 of 4

Now we plug in net force in second Newton’s law:

$$
begin{align*}
F&=ma\
m&=dfrac{F}{a} \
&=dfrac{2.5 mathrm{N}}{6.25 mathrm{m/s^2}} \
&=boxed{0.4 mathrm{kg}}
end{align*}
$$

Result

4 of 4

Mass of a toy is $m=0.4 mathrm{kg}$.

Exercise 26

Solution 1

Solution 2

Step 1

1 of 2

Yes, for a while the diver is accelerating upward because there is an additional upward force due to air resistance on the parachute. The upward acceleration causes the driver’s downward velocity to decrease. Newton’s second law says that a net force in a certain direction will result in an acceleration in that direction (Fnet 􏰅=ma).

Result

2 of 2

Step 1

1 of 3

**Before the parachute opens**

Since the person is falling at a constant speed, we can assume that, until the parachute opens, there is no force on the diver (As then the acceleration is $0$ in Newton’s second law).

Of course, in the real world, a person could never go in a free fall and have a constant velocity, due to Earth’s gravitational force that is always giving the acceleration of $a=-g$. This is just an assumption for this specific problem, where we could approximate the velocity to be constant for a short period of the free fall.

Step 2

2 of 3

**After the parachute opens**

When the parachute opens, there is a drag force due to air molecules hitting the big surface of the parachute. These molecules are technically pushing the parachute up, as from the person’s relative perspective the molecules are hitting the person in the free fall.

This drag force is thus pushing upwards. So the drag force gives upwards acceleration. This creates the effect of initial constant velocity lowering. So if we take the downwards direction to be positive, the acceleration is negative (Downwards deceleration).

Step 3

3 of 3

Note: Since the drag force is proportional to the velocity of the object going through the fluid, as the upwards acceleration decreases the velocity, the drag force comes in equilibrium with the gravitational force, so the velocity again becomes constant due to $0$ net force. This new constant velocity is what keeps the person from smashing into the ground, so the person can land safely!

Exercise 27

Step 1

1 of 2

Since the 1000 N crate has already been weighed, its mass is known. Before loading, the force required to move that crate along the belt should be assessed. Newton’s second law reads: $F = mcdot a$. If we move all the crates acting with the same force as the 1000 N box, we can estimate their weight by observing the acceleration. Boxes with higher acceleration have less weight and those with lower acceleration have more.

Result

2 of 2

The weight of the following crates can be estimated based on the force required for loading.

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Page 101: Section Review

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