Q34E - University Physics with Modern Physics

Question

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. 

(a) What is the force constant of this spring? 

(b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? 

(c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?

Step-by-Step Solution

Answer

a) The force constant of the spring is $2.67\times {10}^{4 N/m}$

b) The force on the spring is $801N$ .

c) The work done to the compression of spring is 21.4 J , and force on the spring to be compressed is 1070 N 

Step 1: Identification of given data

The given data can be listed below,

• The work done on the spring is, $W=12 J$ .
• The initial stretch length of the spring is, $x_1=0 cm$ .
• The compressed length of the spring is, $x_2=3.0 cm$ .

Step 2: Concept/Significance of hook's law

The simplest connection describing the constitutive behaviour of elastic materials is Hooke's law. It is stated that, for generally minor deformations.

Step 3: (a) Determination of the force constant of this spring

The work done on the spring is given by,

 

$W=\frac{1}{2}k{x}_1^2-\frac{1}{2}k{x}_2^2$ 

 

Here, k is the force constant of the spring, $x_1$ is the initial stretch in the spring and $x_2$ is the compressed length of the spring.

 

$$\begin{gathered} 12 J=\frac{1}{2}k{\left(0\right)}^2-\frac{1}{2}k{\left(3 cm\right)}^2 \\ k=\frac{2\times 12 J}{{\left(3 cm\times {\displaystyle \frac{{10}^{-2}m}{1 cm}}\right)}^2} \\ k=2.67\times {10}^{4}N/m \end{gathered}$$

 

Thus, the force constant of the spring is $2.67\times {10}^{4}N/m$ .

Step 4: (b) Determination of the magnitude force is needed to stretch the spring 3.00 cm from its unstretched length

The force on the spring is given by,

 

$F=kx$

 

Here, k is the force constant of the spring, and x is the length of the stretched spring.

 

Substitute all the values in the above,

 

$$\begin{gathered} F=\left(2.67\times {10}^{4}N/m\right)\left(3cm\right)\left(\frac{{10}^{-2}m}{1 cm}\right) \\ =801 N \end{gathered}$$ 

 

Thus, the force on the spring is 801 N .

Step 5: (c) Determination of work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance

 mThe work done to the compression of spring is given by,

 

$W=\frac{1}{2}k{x}_1^2-\frac{1}{2}k{x}_2^2$

 

Here, k is the force constant of the spring, $x_1$ is the initial stretch in the spring and $x_2$ is the compressed length of the spring whose value is 0.040 . 

 

Substitute all the values in the above,

 

$$\begin{gathered} W=\frac{1}{2}\left(2.67\times {10}^{4}N/m\right){\left(-0.040m\right)}^2 \\ =21.4 J \end{gathered}$$

 

The force on the spring to be compressed is given by,

 

$F_x=kx$

 

Here, k is the force constant of the spring, and x is the length of the com spring.

 

Substitute all the values in the above,

 

$$\begin{gathered} F_x=\left(2.67\times {10}^{4}N/m\right)\left(0.040 m\right) \\ =1070 N \end{gathered}$$

 

Thus, the work done to the compression of spring is 21.4 J and force on the spring to be compressed is 1070 N .