(1)You push on a spring whose stiffness is 11 N/m, compressing it until it is 2.5 cm shorter than its relaxed length. What is the magnitude of the force the spring now exerts on your hand?

(2) A different spring is 0.17 m long when it is relaxed.

(a) When a force of magnitude 250 N is applied, the spring becomes 0.24 m long. What is the stiffness of this spring?

(b) This spring is compressed so that length is 0.15 m. What magnitude of the force is required to do this?

The given data can be listed below as,

• The value of the spring stiffness is,$k=11\text{\hspace{0.17em}N/m}$
• The change in length of the first spring is,$x_1=2.5\text{\hspace{0.33em}cm}=2.5\text{\hspace{0.33em}cm}\times \frac{1\text{\hspace{0.33em}m}}{100\text{\hspace{0.33em}cm}}=2.5\times {10}^{-2}\text{\hspace{0.33em}m}$
• The change in length of the second spring is,$x_2=0.17\text{\hspace{0.33em}m}$
• The magnitude of the force is,$F=250\text{\hspace{0.33em}N}$
• After the application of the force, the change in length of the spring is,$x_3=0.24\text{\hspace{0.33em}m}$
• The final change in length is,$x_4=0.15\text{\hspace{0.17em}m}$

The change in distance due to the force required to extend or compress the spring is directly proportional to the distance. It is expressed as follows,

$F=-kx$

Here, -k is the spring constant and x is the change in the distance.

(1)

Substitute all the values in the above expression.

$\begin{array}{c}\left|F\right|=\left|-k{x}_1\right|\\ =\left(11\text{\hspace{0.33em}N/m}\right)\left(2.5\times {10}^{-2}\text{\hspace{0.33em}m}\right)\\ =38.5\times {10}^{-2}\text{\hspace{0.33em}N}\\ =0.385\text{\hspace{0.33em}N}\end{array}$

Thus, the magnitude of the force on the hand is 0.385 N.

(2)

(a)

Determine the net change in length of the second spring.

$\begin{array}{c}x\text{'}=x_3-x_2\\ =0.24\text{\hspace{0.33em}m}-0.17\text{\hspace{0.33em}m}\\ =0.07\text{\hspace{0.33em}m}\end{array}$

Determine the stiffness of the second spring using the mentioned Hooke's law.

$\begin{array}{c}250\text{\hspace{0.33em}N}=k\left(0.07\text{\hspace{0.33em}m}\right)\\ k=\frac{250\text{\hspace{0.33em}N}}{0.07\text{\hspace{0.33em}m}}\\ =3571.4\text{\hspace{0.33em}N/m}\end{array}$

Thus, the stiffness of the second spring is$3571.4\text{\hspace{0.33em}N/m}$

(b)

Determine the net change in length of the second spring.

$\begin{array}{c}x\text{'}=x_2-x_4\\ =0.17\text{\hspace{0.33em}m}-0.15\text{\hspace{0.33em}m}\\ =0.02\text{\hspace{0.33em}m}\end{array}$

Determine the magnitude of the force of the second spring using the mentioned Hooke's law.

$\begin{array}{c}F=\left(3571.4\text{\hspace{0.17em}N/m}\right)\cdot \left(0.02\text{\hspace{0.17em}m}\right)\\ =71.4\text{\hspace{0.33em}N}\end{array}$

Thus, the magnitude of the force of the second spring is 71.4 N