Q90P - University Physics with Modern Physics

Question

Figure P6.90 shows the results of measuring the force F exerted on both ends of a rubber band to stretch it a distance x from its unstretched position. (Source: www.sciencebuddies.org) The data points are well fit by the equation $\mathit{F}\mathbf{=}\mathbf{33}\mathbf{.}\mathbf{55}{\mathit{x}}^{\mathbf{0}\mathbf{.}\mathbf{4571}}$ , where F is in newtons and x is in meters. (a) Does this rubber band obey Hooke’s law over the range of x shown in the graph? Explain. (b) The stiffness of a spring that obeys Hooke’s law is measured by the value of its force constant k, where k = F/x. This can be written as k = dF/dx to emphasize the quantities that are changing. Define ${\mathbf{k}}_{\mathbf{eff}}\mathbf{=}\mathbf{dF}\mathbf{/}\mathbf{dx}$ and calculate k_eff as a function of x for this rubber band. For a spring that obeys Hooke’s law, is constant, independent of x. Does the stiffness of this band, as measured by k_eff , increase or decrease as x is increased, within the range of the data? (c) How much work must be done to stretch the rubber band from x = 0 to x = 0.0400 m ? From x=0.0400 m to x=0.0800 m ? (d) One end of the rubber band is attached to a stationary vertical rod, and the band is stretched horizontally 0.0800 m from its unstretched length. A 0.300-kg object on a horizontal, frictionless surface is attached to the free end of the rubber band and released from rest. What is the speed of the object after it has traveled 0.0400 m?

Step-by-Step Solution

Answer

1. The rubber does not follow Hook’s law.
2. The stiffness of the band will decrease.
3. The work done required to stretch the band from x = 0 to x = 0.0400 is 0.188 J , and work done required to stretch the band from x = 0.0400 to x = 0.0800  is 0.339 J.
4. The speed of the object is 1.5 m/s .

Step 1: Identify the given data

The equation of force is given by,

 $\mathrm{F}=33.55\left({\mathrm{x}}^{0.4871}\right)$

 

The force constant is given by,

$\mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}}$

Step 2: Concept/Significance of Hook’s law

According to the Hook’s law, the magnitude of force for stretched rubber band is given by,

$$\begin{gathered} \mathbf{F}\mathbf{\propto }\mathbf{x} \\ \mathbf{F}\mathbf{=}\mathbf{kx} \end{gathered}$$ 

Here, x is the amount of rubber stretched and k is the restoring force constant that depends on the material.

Step 3: Determine whether the rubber band obeys Hooke’s law over the range of x (a)

From the Hook’s law,

$F\propto x$ 

 

The extension in length of the rubber band is directly proportional to the applied force. So, the graph between F  and x  must be a straight line starting from the origin with some finite positive slope.

 

But from the graph, it can be observed that it does not represent this proportionality.

 

Therefore, the rubber does not follow Hook’s law.

Step 4: Determine whether the stiffness of this band, as measured by k eff , increases or decreases as x is increased x (b)

Consider the given equation.

$\mathrm{F}=33.55\left({\mathrm{x}}^{0.4871}\right)$

 

Simplify for $k_{eff}$ as follows.

$$\begin{gathered} {\mathrm{k}}_{\mathrm{eff}}=\frac{\mathrm{dE}}{\mathrm{dx}} \\ =\frac{\mathrm{d}}{\mathrm{dx}}\left[33.55{\mathrm{x}}^{0.4871}\right] \\ =33.55\left(0.4871\right){\mathrm{x}}^{0.4871-1} \\ =\frac{16.34}{{\mathrm{x}}^{0.5129}} \end{gathered}$$ 

 

From the above relation, it can be observed that ${\mathrm{k}}_{\mathrm{eff}}$ is inversely proportional to x. So, as the value of x increases, ${\mathrm{k}}_{\mathrm{eff}}$ decreases.

 

Therefore, the stiffness of the band will decrease.

Step 5: Determine the work done to stretch the rubber band from x = 0 to x = 0.0400 m and from x = 0.0400 m to x = 0.0800 m (c)

The work done for $x=0$ to $x=0.0400$ is calculated as follows.

$$\begin{gathered} W={\int }_0^{0.0400}Fdx \\ ={\int }_0^{0.0400}\left(33.55x^{0.4871}\right)dx \\ =\frac{\left(33.55\right)}{1.4871}{\left(0.0400\right)}^{1.4871} \\ =0.188 \mathrm{J} \end{gathered}$$ 

 

Therefore, the required work done is 0.188 J .

 

The work done for x = 0.0400 to x = 0.0800 is calculated as follows.

$$\begin{gathered} W={\int }_0^{0.0400}Fdx \\ ={\int }_{0.0400}^{0.0800}\left(33.55x^{0.4871}\right)dx \\ =\frac{\left(33.55\right)}{1.4871}\left[{\left(0.0800\right)}^{1.4871}-{\left(0.0400\right)}^{1.4871}\right] \\ =0.339 \mathrm{J} \end{gathered}$$ 

 

Therefore, the required work done is 0.339 J .

Step 6: Determine the speed of the object after it has traveled 0.0400 m (d)

The work-energy theorem when applies to the object gives,

$$\begin{gathered} W=K_2-K_1 \\ W=\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}^2 \end{gathered}$$

 

Here, W is the work done, K₂ is the final kinetic energy, K₁ is the initial kinetic energy, v₁ is the final speed, v and is the initial speed.

 

Substitute 0.300 kg for m, 0.339 J  for W, and 0 m/s for v in equation (1), we get

 $$\begin{gathered} 0.339 \mathrm{J}=\frac{1}{2}\left(0.339 \mathrm{kg}\right){\mathrm{v}}_1^2-\frac{1}{2}\left(0.300 \mathrm{kg}\right){\left(0 \mathrm{m}/\mathrm{s}\right)}^2 \\ {\mathrm{v}}_1=\sqrt{\frac{2\left(0.339 \mathrm{J}\right)}{\left(0.300 \mathrm{kg}\right)}} \\ {\mathrm{v}}_1=1.5 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Therefore, the speed of the object is 1.5 m/s .