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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 442: Lesson Check

Exercise 49

Step 1

1 of 2

The first one will return to its’ original length while the other will have greater length but it won’t break.

Result

2 of 2

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Exercise 50

Step 1

1 of 2

Hooke’s law says that the elongation or contraction of an object is directly proportional to the magnitude of the force that elongates or contracts it.

Result

2 of 2

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Exercise 51

Step 1

1 of 2

By definition, the elastic object returns to its’ original length when the force that deforms it stops acting.

Result

2 of 2

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Exercise 52

Step 1

1 of 2

The applied force increases or decreases (depending on its’ direction) the distance between the individual atoms and molecules that make the body.

Result

2 of 2

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Exercise 53

Step 1

1 of 2

The resulting spring constant is less than that of individual springs since both of them will be elongated when the force is applied so we have more net elongation applying the same force.

Result

2 of 2

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Exercise 54

Step 1

1 of 3

### Knowns

– The force acting on a spring $F = 5.2 text{ N}$

– The elongation of the spring $x = 14text{ cm}$

Step 2

2 of 3

### Calculation

When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.

We will solve the problem by applying Hooke’s Law :

$$
begin{equation*}
F = k , x
end{equation*}
$$

Rearranging for $k$ we have:

$$
begin{equation*}
k = frac{F}{x}
end{equation*}
$$

Plugging in the numbers we get:

$$
begin{align*}
k = frac{5.2text{ N}}{0.14text{ m}} = 37.14 ; frac{text{N}}{text{m}}
end{align*}
$$

Result

3 of 3

The spring constant is $k = 37.14 ; frac{text{N}}{text{m}}$

Exercise 55

Step 1

1 of 3

### Knowns

– The force acting on a spring $F = 6.1 text{ N}$

– The spring constant of the spring $k = 56 ; frac{text{N}}{text{m}}$

Step 2

2 of 3

### Calculation

When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.

We will solve the problem by applying Hooke’s Law :

$$
begin{equation*}
F = k , x
end{equation*}
$$

Rearranging for $x$ we have:

$$
begin{equation*}
x = frac{F}{k}
end{equation*}
$$

Plugging in the numbers we get:

$$
begin{align*}
x = frac{6.1text{ N}}{56 ; frac{text{N}}{text{m}}} approx 0.109text{ m}
end{align*}
$$

Result

3 of 3

The spring elongation is $x = 10.9text{ cm}$

Exercise 56

Step 1

1 of 3

### Knowns

– The elongation of the spring $x = 19text{ cm}$

– The spring constant of the spring $k = 62 ; frac{text{N}}{text{m}}$

Step 2

2 of 3

### Calculation

When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.

We will solve the problem by applying Hooke’s Law :

$$
begin{equation*}
F = k , x
end{equation*}
$$

Plugging in the numbers we get:

$$
begin{align*}
F = 62 ; frac{text{N}}{text{m}} cdot 0.19text{ m} approx 11.8text{ N}
end{align*}
$$

Result

3 of 3

The applied force is $F = 11.8text{ N}$

Exercise 57

Step 1

1 of 2

We know that the elongations of individual springs are

$$
x_1 = frac{K}{k_1},quad x_2 =frac{F}{k_2}.
$$

The total elongation is

$$
x=x_1+x_2 = frac{F}{k_1}+frac{F}{k_2} = Ffrac{k_1+k_2}{k_1k_2}
$$
which can be rewritten as

$$
F= frac{k_1k_2}{k_1+k_2}x
$$

so this system acts as a spring of constant

$$
k=frac{k_1k_2}{k_1+k_2} = 26text{ N/m}.
$$

Result

2 of 2

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