(1) A spring of stiffness 13 N/m, with relaxed length 20 cm, stands vertically on a table as shown in Figure 2.36. Use the usual coordinate system, with +x to the right, +y up, and +z out of the page, towards you. (a) When the spring is compressed to a length of 13 cm, what is the unit vector $\hat{\mathbf{L}}$? (b) When the spring is stretched to a length of 24 cm, what is the unit vector $\hat{\mathbf{L}}$? (2) A different spring of stiffness 95 N/m, and with relaxed length 15 cm, stands vertically on a table, as shown in Figure 2.36. With your hand you push straight down on the spring until your hand is only 11 cm above the table. Find (a) the vector $\hat{\mathbf{L}}$, (b) the magnitude of $\hat{\mathbf{L}}$, (c) the unit vector role="math" localid="1668490124469" $\hat{\mathbf{L}}$, (d) the stretch s, (e) the forcerole="math" localid="1668490004012" $\overrightarrow{\mathbf{F}}$ exerted on your hand by the spring.

The given data can be listed below as,

(1)

The stiffness of the spring is, k = 13 N/m

The relaxed length of the spring is, $\mathrm{x}=20\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.2\mathrm{m}$

(2)

The stiffness of the spring is, k = 95 N/m

The relaxed length of the spring is,$\mathrm{x}=15\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.15\mathrm{m}$

The force exerted by the spring can be determined by taking the product of spring stiffness and the change in length when the spring is compressed or stretched. It is expressed as follows,

F = -kx ....(1)

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

The unit vector can be determined by dividing the vector by its magnitude. It is expressed as follows,

$\hat{x}=\frac{\overrightarrow{x}}{\left|\overrightarrow{x}\right|}...\left(2\right)$

Here, $\overrightarrow{x}$is the vector.

(1)

(a)

The length to which the spring is compressed is $\mathrm{L}=13\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.13\mathrm{m}$.

Write the expression for the vector form of length if it is extended from the fixed to the moved point.

Determine the magnitude of the above vector.

$$\begin{gathered} \left|\overrightarrow{L}\right|=\sqrt{0^2+0.{13}^2+0^2} \\ =0.13\mathrm{m} \end{gathered}$$

Determine the unit vector for compressed length using equation (2).

Thus, the required unit vector is

(b)

The length to which the spring is stretched is $L=24\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.24\mathrm{m}$.

Write the expression for the vector form of length if it is extended from the fixed to the moved point.

Determine the magnitude of the above vector.

$$\begin{gathered} \left|\overrightarrow{L}\right|=\sqrt{0^2+0.{24}^2+0^2} \\ =0.24\mathrm{m} \end{gathered}$$

Determine the unit vector for compressed length using equation (2).

Thus, the required unit vector is and negative sign indicates the length of the spring is larger than that of the relaxed length.

(2)

(a)

The length to which the spring is compressed is $L=11\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.11\mathrm{m}$.

Write the expression for the vector form of length if it is extended from the fixed to the moved point.

Thus, the required vector is .

(b)

Determine the magnitude of the above vector.

$$\begin{gathered} \left|\overrightarrow{L}\right|=\sqrt{0^2+0.{11}^2+0^2} \\ =0.11\mathrm{m} \end{gathered}$$

Thus, the required magnitude is 0.11 m.

(c)

Determine the unit vector for compressed length using equation (2).

(d)

Determine the spring stretch by subtracting relaxed length of the spring from the magnitude of the vector of length when the spring is compressed.

$$\begin{gathered} s=\left|\overrightarrow{L}\right|-x \\ =0.11-0.15 \\ =0.04\mathrm{m} \end{gathered}$$

Thus, the spring stretch when spring is compressed is 0.04 m.

(d)

Determine the force applied on the hand by the spring using equation (1).

Thus, the force applied on the hand by the spring is .