Q6-38E - University Physics with Modern Physics

Question

 A spring of force constant $\mathbf{300}\mathbf{.0}\mathbf{\text{\hspace{0.17em}N/m}}$and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

Step-by-Step Solution

Answer

The stretch length of the spring is $0.050\text{\hspace{0.17em}m}$and work done to stretch the spring is$0.375\text{\hspace{0.17em}J}$

 

Step 1: Identification of given data

The given data can be listed below,

• The force constant of spring is,$k=300\text{\hspace{0.17em}N/m}$.
• The initial unstretched length of the spring is$l_0=0.240\text{\hspace{0.17em}m}$,.
• The force applied in the opposite direction on the spring is, $F=15\text{\hspace{0.17em}N}$.

 

Step 2: Concept/Significance of hook’s law

According to the dynamical principle of hooks law, an elastic spring will exert a force proportional to how much it has been compressed or extended from its equilibrium length. It only holds true for sufficiently tiny spring stretching or compression.

Step 3: Determination of the length of the spring after stretching and work done required to stretch the spring

The stretched amount of length of the spring is given by,

$\begin{array}{c}F=kx\\ x=\frac{F}{k}\end{array}$ 

 

Here, F is the force on the spring, and k is the force constant.

 

Substitute all the values in the above,

 

$\begin{array}{c}\mathrm{x}=\frac{15\text{\hspace{0.17em}N}}{300\text{\hspace{0.17em}N/m}}\\ =0.05\text{\hspace{0.17em}m}\end{array}$

 

The total length of spring after stretching is given by,

$\mathrm{l}=\mathrm{x}+{\mathrm{l}}_0$ 

 

Here, x is the stretched length and is the initial length of the spring.

 

Substitute all the values in the above,

 $\begin{array}{c}\mathrm{l}=(0.05+0.240)\text{\hspace{0.17em}m}\\ =0.290\text{\hspace{0.17em}m}\end{array}$

 

The work required to stretch the spring is given by,

 $W=\frac{1}{2}k{x}^2$

 

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

 

Substitute all the values in the above,

 

$\begin{array}{c}W=\frac{1}{2}\left(300\text{\hspace{0.17em}N/m}\right){\left(0.05\text{\hspace{0.17em}m}\right)}^2\\ =\left(150\text{\hspace{0.17em}N/m}\right)\left(0.0025{\text{\hspace{0.17em}m}}^2\right)\\ =0.375\text{\hspace{0.17em}N}\cdot \text{m}\left(\frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}}\right)\\ =0.375\text{\hspace{0.17em}J}\end{array}$

 

Thus, the stretch length of the spring is and the work done to stretch the spring is$0.375\text{\hspace{0.17em}J}$ .