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 ball leaves the barrel if you can ignore friction.

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wagonbelleville wagonbelleville · Answer 1 · 2019-11-19T08:06:05+01:00

Answer:

v = 6.93 m/s

Explanation:

given,

spring constant = k = 400 N/m

compression of spring = 6 cm = 0.06 m

mass of ball = 0.03 Kg

length of barrel = 6 cm

using energy conservation

KE of ball = PE of spring

[tex]\dfrac{1}{2}mv^2 =\dfrac{1}{2}kx^2[/tex]

[tex]mv^2 = kx^2[/tex]

[tex]v = \sqrt{\dfrac{kx^2}{m}}[/tex]

[tex]v = \sqrt{\dfrac{400\times 0.06^2}{0.03}}[/tex]

[tex]v = \sqrt{48}[/tex]

v = 6.93 m/s

hence, the speed at which ball leaves the barrel will be equal to v = 6.93 m/s