Q29P - Fundamentals Of Physics

Question

In the arrangement of Fig. $\mathbf{7}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{10}$, we gradually pull the block from $\mathit{x}\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathit{t}\mathit{o}\mathbf{ }\mathbf{ }\mathit{x}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{c}\mathit{m}$ , where it is stationary. Figure  $7-35$gives the work that our force does on the block. The scale of the figure’s vertical axis is set by   ${\mathit{W}}_{\mathbf{s}}\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{ }\mathit{J}$ . We then pull the block out to  $\mathit{x}\mathbf{=}\mathbf{+}\mathbf{ }\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{ }\mathit{c}\mathit{m}\mathbf{ }$ and release it from rest. How much work does the spring do on the block when the block moves from${\mathit{x}}_{\mathbf{i}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{c}\mathit{m}$  to, (a) $\mathit{x}\mathbf{=}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{c}\mathit{m}$ (b) $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{ }\mathit{c}\mathit{m}$, and (c) $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{ }\mathit{c}\mathit{m}$?

Step-by-Step Solution

Answer

1. Work done is, $W_{\text{s}}=0.90 \text{J}$.
2. Work done is,$W_{\text{s}}=2.1 \text{J}$ .
3. Work done is,$W_{\text{s}}=0 \text{J}$ .

 

Step 1: Given data

1. The initial position of the block is $x_{\text{i}}=0 \text{m}$ .
2. The vertical axis of the graph shows work done by the spring on the block as, $W_{\text{s}}=1.0 \text{J}$.

Step 2: Understanding the concept

The block is connected to the spring; hence, we can use the concept of the work by the spring on the block, which is associated with the change in energy.

 Formulae:

 

   $W_{\text{s}}=\frac{1}{2}k{x}_{\text{i}}^{\text{2}}-\frac{1}{2}k{x}_{\text{f}}^{\text{2}}$                                                                                                                     (1)

Here  $x_{\text{f}}$ is the final position of the block and  is the spring constant. 

Step 3: Calculate the value of k

 From the graph, 

 

For the vertical axis of${\text{x}}_{\text{i}}=2 \text{cm}=0.020 \text{m}$ , work done is .

$x_{\text{f}}$  will be 0.

 

The work done by the spring can be calculated as,

 $$\begin{gathered} W_{\text{s}}=\frac{1}{2}k\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right) \\ k=\frac{2W_s}{\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)} \end{gathered}$$

Substitute the values and find the value of spring constant as

$$\begin{gathered} k=\frac{2\times 0.40 \text{J}}{\left[{\left(0.020 \text{m}\right)}^2-0^2\right]} \\ k=2\times {10}^3 \text{N/m} \end{gathered}$$

Step 4: (a) Calculate the work the spring does on the block when the block moves from x i = + 5 . 0   c m to , x f = + 4 . 0   c m

When the block moves from$x_{\text{i}}=+5.0 \text{cm}=0.050 \text{m}$   to $x_{\text{f}}=4.0 \text{cm}=0.040 \text{m}$ , the work by the spring can be calculated using equation 1 as,

 $$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3 \text{N/m}\times \left[{\left(0.050 \text{m}\right)}^2-{\left(-0.040 \text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3 \text{N/m}\times \text{9}\times {\text{10}}^{-4} {\text{m}}^2 \\ {W}_{\text{s}}=0.90 \text{J} \end{gathered}$$

Thus, the work done is, $W_{\text{s}}=0.90 \text{J}$.

Step 5: (b) Calculate the work the spring does on the block when the block moves from x i = + 5 . 0   c m to x f = - 2 . 0     c m

When the block moves from   $x_{\text{i}}=+5.0 \text{cm}=0.050 \text{m}$to $x_{\text{f}}=-2.0 \text{cm}=-0.020 \text{m}$ , the work done by the spring can be calculated using equation 1 as,

 $$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3 \text{N/m}\times \left[{\left(0.050 \text{m}\right)}^2-{\left(-0.020 \text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3 \text{N/m}\times \text{2.1}\times {\text{10}}^{-3} {\text{m}}^2 \\ {W}_{\text{s}}=2.1 \text{J} \end{gathered}$$

Thus, the work done is, $W_{\text{s}}=2.1 \text{J}$.

 

Step 6: (c) Calculate the work the spring does on the block when the block moves from x i = + 5 . 0   c m to , x f = - 5 . 0     c m

When the block moves from  $x_{\text{i}}=+5.0 \text{cm}=0.050 \text{m}$ to$x_{\text{f}}=-5.0 \text{cm}=-0.050 \text{m}$  , the work by the spring can be calculated using equation 1 as,

 $$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3 \text{N/m}\times \left[{\left(0.050 \text{m}\right)}^2-{\left(-0.050 \text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3 \text{N/m}\times 0 \\ {W}_{\text{s}}=0 \text{J} \end{gathered}$$

Thus, the work done is, $W_{\text{s}}=0 \text{J}$.