32P - Fundamentals Of Physics

Question

Figure 7-37 gives spring force F_x versus position x for the spring–block arrangement of Fig . 7-10 . The scale is set by ${\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{160}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{N}$. We release the block at $\mathbf{x}\mathbf{=}\mathbf{12}\mathbf{ }\mathbf{cm}$. How much work does the spring do on the block when the block moves from ${\mathbf{x}}_{\mathbf{i}}\mathbf{=}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}$ to (a) x= +5.0 cm, (b) x=-5.0 cm, (c) x=-8.0 cm, and (d) x=-10.0 cm ?

Step-by-Step Solution

Answer

1. The work done by the spring on the block when the block moves from ${\mathrm{x}}_{\mathrm{i}}=+8.0 \mathrm{cm} \mathrm{to} {\mathrm{x}}_{\mathrm{f}}=5.0 \mathrm{cm}$ is  $\mathrm{W}=16 \mathrm{J}$
2. The work done by the spring on the block when the block moves from ${\mathrm{x}}_{\mathrm{i}}=+8.0 \mathrm{cm} \mathrm{to} {\mathrm{x}}_{\mathrm{f}}=-5.0 \mathrm{cm}$ is $\mathrm{W}=16 \mathrm{J}$ 
3. The work done by the spring on the block when the block moves from ${\mathrm{x}}_{\mathrm{i}}=+8.0 \mathrm{cm} \mathrm{to} {\mathrm{x}}_{\mathrm{f}}=-8.0 \mathrm{cm}$ is $\mathrm{W}=0 \mathrm{J}$ 
4. The work done by the spring on the block when the block moves from ${\mathrm{x}}_{\mathrm{i}}=+8.0 \mathrm{cm} \mathrm{to} {\mathrm{x}}_{\mathrm{f}}=-10.0 \mathrm{cm}$ is $\mathrm{W}=-14 \mathrm{J}$ 

Step 1: Understand the concept

We can use the equation of Hook’s law to calculate the spring constant from the graph, hence, we can calculate the work done in each case using the equation of work done by the spring.

 

Formulae:

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{-}\mathbf{kx} \\ \mathbf{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{k}\left(x_i^2-x_f^2\right) \end{gathered}$$ 

 

Given:

1. The graph of ${\mathrm{F}}_{\mathrm{x}} \mathrm{vs} \mathrm{x}$ 
2. The scale of graph is set at, ${\mathrm{F}}_{\mathrm{s}}=160.0 \mathrm{N}$ 

Step 2: Calculate the value of spring constant

The graph of F_x vs x is given, so according to Hook’s law, the slope of this graph will give us the spring constant of the spring used.

From this graph, we get 

$\begin{array}{rcl}\mathrm{slope}& =& \frac{∆\mathrm{y}}{∆\mathrm{x}}\\ & =& \frac{80 \mathrm{N}}{-1 \mathrm{cm}}\\ & =& -80 \frac{\mathrm{N}}{\mathrm{cm}}\\ & =& -8\times {10}^3 \mathrm{N}/\mathrm{m}.\end{array}$ 

As per Hook’s law, we have

$\mathrm{F}=-\mathrm{kx}$ 

So,
$\mathrm{k}=-\frac{\mathrm{F}}{\mathrm{x}}$ 

Hence, the spring constant becomes,

$\mathrm{k}=8\times {10}^3 \mathrm{N}/\mathrm{m}.$

Step 3: a) Calculate the work does the spring do on the block when the block moves from x i =+8.0 cm t o x f =5.0 cm

The equation for work done by the spring is,

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$ 

 

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08 \mathrm{m} \mathrm{and} {\mathrm{x}}_{\mathrm{f}}=0.05 \mathrm{m}$ . 

So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080 \mathrm{m}\right)}^2-{\left(0.050 \mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& 15.6 \mathrm{J}\\ & \approx & 16 \mathrm{J}\end{array}$ 

 

Therefore, work done is, 16 J. 

Step 4: b) Calculate the work does the spring do on the block when the block moves from x i =+8.0 cm to x f =-5.0 cm

The equation for work done by spring is

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$ 

 

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08 \mathrm{m} \mathrm{and} {\mathrm{x}}_{\mathrm{f}}=-0.05 \mathrm{m}$. So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080 \mathrm{m}\right)}^2-{\left(-0.050 \mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& 15.6 \mathrm{J}\\ & \approx & 16 \mathrm{J}\end{array}$ 

 

Therefore, work done is, 16 J.

Step 5: c) Calculate the work does the spring do on the block when the block moves from x i =+8.0 cm to x f =-8.0 cm

The equation for work done by spring is

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$ 

 

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08 \mathrm{m} \mathrm{and} {\mathrm{x}}_{\mathrm{f}}=-0.08 \mathrm{m}$. So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080 \mathrm{m}\right)}^2-{\left(-0.080 \mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& 0 \mathrm{J}\end{array}$ 

 

Therefore, work done is, 0 J.

Step 6: d) Calculate the work does the spring do on the block when the block moves from x i =+8.0 cm to x f =-10.0 cm

We have, the equation for work done by spring as, 

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$ 

 

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08 \mathrm{m} \mathrm{and} {\mathrm{x}}_{\mathrm{f}}=-0.10 \mathrm{m}$. So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080 \mathrm{m}\right)}^2-{\left(-0.10 \mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& -14.4 \mathrm{J}\\ & \approx & -14 \mathrm{J}\end{array}$ 

 

Therefore, work done is, -14 J.