Question

I got 6 m/s.

The spring of a spring gun has force constant k=400N/m and negligible mass. The spring is compressed 0.06m and a ball with mass 0.03kg 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 0.06m 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 of the ball as it leaves the barrel if a constant resisting force of 6.00N acts on the ball as it moves along the barrel.

Answer 1

\[P.E_{spring}=K.E_{ball}-Work\ done\ by\ resisting\ force\]
\[P.E_{spring}=k\times x\]\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ =(400N/m\times0.06m)\ m\]\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ =24N\]
\[K.E_{ball}\ \ \ =1/2\times m_{ball}\times v^2_{ball}\]\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ =1/2\times 0.03\times v^2_{ball}\]\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ =(0.015\times v^2_{ball})N\]
\[Work\ done\ by\ resisting\ force\ =\ F_{resistance}\times d_{barrel}\]\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 6\times 0.06\]\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 0.36N\]
Therefore\[24=(0.015\times v^2_{ball})-0.36\]\[v_{ball}=\sqrt{24.36/0.015}\]\[\ \ \ \ \ \ =\sqrt{24360/15}\]\[\ \ \ \ \ \ =\sqrt{8120/5}\]\[\ \ \ \ \ \ =\sqrt{1624}\]\[v_{ball}=12.7279\ m/s\]

Answer 2

sorry the ans was 40.2989 m/s

Answer 3

according to me answer should be 4.89 m/s