# To express : Hooke’s law for spring as equation.

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.11, Problem 29E

(a)

To determine

## To express: Hooke’s law for spring as equation.

Expert Solution

Hooke’s law for spring is F=kx .

### Explanation of Solution

Hooke’s law:

The force required to keep the spring stretched beyond the natural length is directly proportional to the extension of spring beyond natural length.

F is directly proportional to x

Mathematically,

F=kx

Here,

F is force needed to keep the spring stretched.

k is spring constant.

x is the extension of spring beyond natural length.

Thus, Hooke’s law as an equation is F=kx .

(b)

To determine

### To find: The spring constant k .

Expert Solution

The value of spring constant is 7.5Ncm .

### Explanation of Solution

Given:

Natural length of spring ( x0 ) is 5cm .

Force required to stretch it up to 9cm is 30N .

Calculation:

The spring is stretched up to 9cm , as the natural length is 5cm , it is stretched 4cm beyond the natural length. So, the value of x is 4cm .

Use Hooke’s Law,

F=kx

Substitute, 30 for F and 4 for x in Hooke’s law equation and solve for k .

30=k4

Multiply both the sides by (14)

(14)30=(14)k47.5=k (1)

Thus, the value of k is 7.5Ncm .

(c)

To determine

### To find: The force required to keep the spring stretched to 11 cm .

Expert Solution

The force required is 45N .

### Explanation of Solution

Given:

The stretched length of the spring is 11cm .

Calculation:

The natural length of spring is 5cm , but spring is stretched to a length of 11cm i.e. 6cm beyond the natural length. So, the value of x is 6cm .

Use Hooke’s Law,

F=kx

Substitute 6 for x and 7.5 for k from equation (1) and solve for F .

F=(7.5)(6)F=45

Thus, the value of F is 45N .

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