Question

Consider the popgun in Example 8.3. Suppose the projectile mass, compression distance, and spring constant remain the same as given or calculated in the example. Suppose, however, there is a friction force of magnitude 2.00 N acting on the projectile as it rubs against the interior of the barrel. The vertical length from point A to the end of the barrel is 0.600 m. (a) After the spring is compressed and the popgun fired, to what height does the projectile rise above point B? (b) Draw four energy bar charts for this situation, analogous to those in Figures 8.6c–d.

Short answer

(a)The height to which the projectile rises above point B is 16.5 m.

(b) The bar charts are drawn below.

Step-by-step solution

Introduction

A force acting between two surfaces in contact, which are in relative motion with each other and act in the direction opposite to that of that motion, is called friction. If a friction force of magnitude acts over a distanced within a system, the change in the internal energy of the system is mathematically presented as:

$\mathbf{∆}{\mathit{E}}_{\mathbf{i}\mathbf{n}\mathbf{t}}\mathbf{=}{\mathit{f}}_{\mathbf{k}}\mathit{D}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here:

$f_k=$Frictional force

d= Distance in meter

$∆E_{int}=$Change in internal energy

Given Data

In Example 8.3:

The mass of the projectile:$m=35.0g$

Compression distance:$y_A=-0.120m$

Force constant:$k=958N/m$

Length of the barrel:$d=0.600m$

Frictional force:$f_k=2.00N$

Position of point B:$y_B=0$

Calculation for part (a)

For the ball spring-Earth system,

$k=0,U_A=mg{y}_A,U_s=\frac{1}{2}k{x}^{2}wherex=\left|y_A\right|;k_f=0;U_C=mg{y}_{C}and{U}_B=0$

$\begin{array}{c}\Delta K+\Delta U+\Delta E_{\mathrm{int}}=0\mathrm{..........}\left(\because \Delta E_{\mathrm{int}}=f_{k}d\right)\\ \Delta K+\Delta U=-f_{k}d\\ 0+\left(mg{y}_C-mg{y}_A\right)+\left(0-\frac{1}{2}k{x}^2\right)=-f_{k}d\\ mg{y}_C=mg{y}_A+\frac{1}{2}k{x}^2-f_{k}d\end{array}$

Solve further

$\begin{array}{c}{y}_C=y_A+\frac{\frac{1}{2}k{x}^2}{mg}\\ =-0.120\text{m}+\frac{\frac{1}{2}\left(958\text{N/m}\right){\left(0.120\text{m}\right)}^2-\left(2.00\text{N}\right)\left(0.600\text{m}\right)}{\left(0.035\text{kg}\right)g}\\ =16.5\text{m.}\end{array}$

The height above point B to which the projectile rises is $y_c=16.5m$

Bar Charts

The energy bar charts that represent the condition given in the equation are drawn below.