Hector Arias

2023-03-30

A spring gun's spring has a constant force k =400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring.The ball is then launched out of the gun's barrel after the spring is released. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. Calculate the speed with which the ballleaves the barrel if you can ignore friction. Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel. For the situation in part (b), at what position along the barrel does the ball have the greatest speed?

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We'll use the principle of conservation of mechanical energy. The total mechanical energy of the ball-spring system will be conserved throughout the motion, neglecting friction.
Let's begin by finding the potential energy stored in the compressed spring. The potential energy stored in a spring can be calculated using the formula:
$PE=\frac{1}{2}k{x}^{2}$
where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.
Given:
Spring constant, $k=400\phantom{\rule{0.167em}{0ex}}\text{N/m}$
Compression, $x=6.00\phantom{\rule{0.167em}{0ex}}\text{cm}=0.06\phantom{\rule{0.167em}{0ex}}\text{m}$
Plugging in the values, we can calculate the potential energy stored in the spring:
$PE=\frac{1}{2}×400\phantom{\rule{0.167em}{0ex}}\text{N/m}×\left(0.06\phantom{\rule{0.167em}{0ex}}\text{m}{\right)}^{2}$
Calculating this, we find:
$PE=0.72\phantom{\rule{0.167em}{0ex}}\text{J}$
Next, we'll determine the kinetic energy of the ball as it leaves the barrel. Since the barrel is horizontal, the potential energy stored in the spring will be fully converted into the kinetic energy of the ball.
The kinetic energy of an object can be calculated using the formula:
$KE=\frac{1}{2}m{v}^{2}$
where $m$ is the mass of the object and $v$ is its velocity.
Given:
Mass of the ball, $m=0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}$
Using the conservation of mechanical energy, we equate the potential energy stored in the spring to the kinetic energy of the ball:
$PE=KE$
$0.72\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}×0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}×{v}^{2}$
Simplifying the equation, we can solve for the velocity $v$:
${v}^{2}=\frac{2×0.72\phantom{\rule{0.167em}{0ex}}\text{J}}{0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}}$
${v}^{2}=48\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$
$v=\sqrt{48\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}$
$v=6.93\phantom{\rule{0.167em}{0ex}}\text{m/s}$
So, the speed with which the ball leaves the barrel, neglecting friction, is 6.93 m/s.
Now, let's move on to part (b) of the problem, where a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. In this case, we need to account for the work done by the resisting force.
The work done by a force is given by the formula:
$W=Fd$
where $W$ is the work done, $F$ is the force, and $d$ is the displacement.
Given:
Resisting force, $F=6.00\phantom{\rule{0.167em}{0ex}}\text{N}$
Barrel length, $d=6.00\phantom{\rule{0.167em}{0ex}}\text{cm}=0.06\phantom{\rule{0.167em}{0ex}}\text{m}$
Plugging in the values, we can calculate the work done by the resisting force:
$W=6.00\phantom{\rule{0.167em}{0ex}}\text{N}×0.06\phantom{\rule{0.167em}{0ex}}\text{m}$
$W=0.36\phantom{\rule{0.167em}{0ex}}\text{J}$
Since work is equal to the change in kinetic energy, we subtract the work done by the resisting force from the initial kinetic energy to find the final kinetic energy of the ball as it leaves the barrel.
Initial kinetic energy of the ball (from part (a)) is 0.72 J.
Final kinetic energy = Initial kinetic energy - Work done by resisting force
$K{E}_{\text{final}}=0.72\phantom{\rule{0.167em}{0ex}}\text{J}-0.36\phantom{\rule{0.167em}{0ex}}\text{J}$
$K{E}_{\text{final}}=0.36\phantom{\rule{0.167em}{0ex}}\text{J}$
Now, we can calculate the final velocity ${v}_{\text{final}}$ using the formula for kinetic energy:
$K{E}_{\text{final}}=\frac{1}{2}m{v}_{\text{final}}^{2}$
Plugging in the values, we can solve for ${v}_{\text{final}}$:
$0.36\phantom{\rule{0.167em}{0ex}}\text{J}=\frac{1}{2}×0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}×\left({v}_{\text{final}}{\right)}^{2}$
$\left({v}_{\text{final}}{\right)}^{2}=\frac{2×0.36\phantom{\rule{0.167em}{0ex}}\text{J}}{0.0300\phantom{\rule{0.167em}{0ex}}\text{kg}}$
$\left({v}_{\text{final}}{\right)}^{2}=24\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}$
${v}_{\text{final}}=\sqrt{24\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}$
${v}_{\text{final}}=4.90\phantom{\rule{0.167em}{0ex}}\text{m/s}$
So, the speed of the ball as it leaves the barrel with a constant resisting force of 6.00 N is 4.90 m/s.
Finally, to determine the position along the barrel where the ball has the greatest speed, we need to consider the work done by the resisting force. The work done is given by the formula:
$W=Fd$
Since the work done is constant, the magnitude of the force multiplied by the displacement will be the same at all positions along the barrel. Therefore, the ball will have the greatest speed at the position where it has traveled the greatest distance.
Since the barrel length is 6.00 cm, the ball will have the greatest speed when it reaches the end of the barrel, at a position of 6.00 cm from the starting point.
Thus, the ball has the greatest speed at the end of the barrel.
To summarize:
- The speed with which the ball leaves the barrel, neglecting friction, is 6.93 m/s.
- The speed of the ball as it leaves the barrel with a constant resisting force of 6.00 N is 4.90 m/s.
- The ball has the greatest speed at the end of the barrel (6.00 cm from the starting point).
Please let me know if anything is unclear or if you have any further questions!

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