Question

A spring has an unstretched length of 10 cm . It exerts a restoring force F when stretched to a length of 11 cm .

Answer 1

Given:
L = 10 cm, original length

Because the stretched length is 11 cm, the extension is
d = 11 - 10 = 1 cm

Let the spring constant be k N/cm
Then the restoring force is
F = (k N/cm)*(1 cm)
   = k N

Answer:
The restoring force is equal to the spring constant, measured in Newtons per centimeter.

Answer 2

(a). The restoring force in the spring will be $3F$  if it is stretched to a length of $\boxed{13\,{\text{cm}}}$ .

(b). The restoring force in the spring will be $2F$  if it is compressed to a length of $\boxed{8\,{\text{cm}}}$ .

Further Explanation:

When we compress or stretch a spring from its natural length, there is a restoring force developed in the spring due to the compression and stretching of the spring.

The restoring force experienced by the spring due to stretching is expressed as:

$F=k\cdot\Delta x$                                                           …… (1)

Here, $F$  is the restoring force developed in the spring, $k$  is the spring constant of the spring and $\Delta x$  is the length through which the spring is stretched.

The spring experiences a restoring force of  $F$ when it is stretched to a length of $11\,{\text{cm}}$  from its natural length $10\,{\text{cm}}$ .

$\begin{aligned}\Delta x&={x_f} - {x_i}\\&=0.10 - 0.11\\&=0.01\,{\text{m}}\\\end{aligned}$

Substitute the values of force and change in length in equation (1).

$\begin{aligned}F&=k\cdot0.01\hfill\\k&=\frac{F}{{0.01}}\hfill\\\end{aligned}$

Part (a):

When the spring experiences a restoring force of $3F$ , then the stretched length of the spring should be:

$3F=k.\Delta x$

Substitute $\frac{F}{{0.01}}$  for $k$  in above expression.

$\begin{aligned}3F&=\frac{F}{{0.01}}\cdot\Delta x' \\\Delta x'&=3\times0.01\,{\text{m}}\\&=3\,{\text{cm}}\\\end{aligned}$

So, the stretched length of the spring becomes:

$\begin{aligned}L&={x_o}+\Delta x' \\&=10\,{\text{cm}}+3\,{\text{cm}}\\&=13\,{\text{cm}}\\\end{aligned}$

Thus, the restoring force in the spring will be $3F$  if it is stretched to a length of $\boxed{13\,{\text{cm}}}$ .

Part (b):

The restoring force of magnitude $2F$  is experienced by the spring on compression. The change in length due to compression will be:

$2F=k\cdot\Delta x''$

Substitute $\frac{F}{{0.01}}$  for  $k$ in above expression.

$\begin{aligned}2F&=\frac{F}{{0.01}}\cdot\Delta x''\\\Delta x''&=2\times0.01\,{\text{m}}\\&=2\,{\text{cm}}\\\end{aligned}$

So, the compressed length of the spring becomes:

$\begin{aligned}L'&={x_o}-\Delta x''\\&=10\,{\text{cm}}-{\text{2}}\,{\text{cm}}\\&=8\,{\text{cm}}\\\end{aligned}$

Thus, the restoring force in the spring will be $2F$  if it is compressed to a length of $\boxed{8\,{\text{cm}}}$ .

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Work and energy

Keywords:

Spring, unstretched length, compressed, stretched, restoring force, 3F, 11 cm, F=kx, natural length of spring.