# The spring of a spring gun has force constant 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 spring is then released, and the ball is propelled out the barrel of the gun. 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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The spring of a spring gun has force constant 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 spring is then released, and the ball is propelled out the barrel of the gun. 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?

2021-02-23
a) $$\displaystyle{\frac{{{1}}}{{{2}}}}{k}{x}^{{2}}={\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}$$
$$\displaystyle{v}=\sqrt{{{\frac{{{k}{x}^{{2}}}}{{{m}}}}}}=\sqrt{{{\frac{{{\left({400}\right)}{\left({0.06}\right)}^{{2}}}}{{{0.03}}}}}}={6.93}\ {\frac{{{m}}}{{{s}}}}$$
b) $$\displaystyle{\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}={W}_{{\text{total}}}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}={\frac{{{1}}}{{{2}}}}{k}{x}^{{2}}-{F}_{{{\mathfrak{{i}}}{c}}}{x}$$
$$\displaystyle{v}=\sqrt{{{\frac{{{k}{x}^{{2}}-{2}{F}_{{{\mathfrak{{i}}}{c}}}{x}}}{{{m}}}}}}$$
$$\displaystyle{v}=\sqrt{{{\frac{{{\left({400}\right)}{\left({0.06}\right)}^{{2}}-{2}{\left({6}\right)}{\left({0.06}\right)}}}{{{0.03}}}}}}={4.90}\ {\frac{{{m}}}{{{s}}}}$$
c) The greatest speed occurs when the acceleration (and the net force)are zero
$$\displaystyle{F}={k}{x}$$
$$\displaystyle{x}={\frac{{{F}}}{{{k}}}}={\frac{{{6}}}{{{400}}}}={0.015}{m}$$
Then to find the speed:
$$\displaystyle{W}_{{\text{total}}}={\frac{{{1}}}{{{2}}}}{k}{\left({{x}_{{0}}^{{2}}}-{x}^{{2}}\right)}-{f{{\left({x}_{{0}}-{x}\right)}}}$$
Solve for velocity;
$$\displaystyle{v}={5.20}\ {\frac{{{m}}}{{{s}}}}$$

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