91影视

The given data can be listed below as,

• The value of the spring stiffness is,$k=11\text{鈥塏/尘}$
• The change in length of the first spring is,$x_1=2.5\text{鈥刢尘}=2.5\text{鈥刢尘}脳\frac{1\text{鈥刴}}{100\text{鈥刢尘}}=2.5脳{10}^{鈭�2}\text{鈥刴}$
• The change in length of the second spring is,$x_2=0.17\text{鈥刴}$
• The magnitude of the force is,$F=250\text{鈥凬}$
• After the application of the force, the change in length of the spring is,$x_3=0.24\text{鈥刴}$
• The final change in length is,$x_4=0.15\text{鈥尘}$

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.

$\begin{array}{c}\left|F\right|=\left|鈭�k{x}_1\right|\\ =\left(11\text{鈥凬/尘}\right)\left(2.5脳{10}^{鈭�2}\text{鈥刴}\right)\\ =38.5脳{10}^{鈭�2}\text{鈥凬}\\ =0.385\text{鈥凬}\end{array}$

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.

$\begin{array}{c}x\text{'}=x_3鈭�x_2\\ =0.24\text{鈥刴}鈭�0.17\text{鈥刴}\\ =0.07\text{鈥刴}\end{array}$

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

$\begin{array}{c}250\text{鈥凬}=k\left(0.07\text{鈥刴}\right)\\ k=\frac{250\text{鈥凬}}{0.07\text{鈥刴}}\\ =3571.4\text{鈥凬/尘}\end{array}$

Thus, the stiffness of the second spring is$3571.4\text{鈥凬/尘}$

(b)

Determine the net change in length of the second spring.

$\begin{array}{c}x\text{'}=x_2鈭�x_4\\ =0.17\text{鈥刴}鈭�0.15\text{鈥刴}\\ =0.02\text{鈥刴}\end{array}$

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

$\begin{array}{c}F=\left(3571.4\text{鈥塏/尘}\right)鈰�\left(0.02\text{鈥尘}\right)\\ =71.4\text{鈥凬}\end{array}$

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