The spring of a spring gun has force constant k = 450 N/m and negligible mass. T

Question

The spring of a spring gun has force constant k = 450 N/m and negligible mass. The spring is compressed 6.40 cm and a ball with mass 0.0270 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 6.40 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.

a) Calculate the speed with which the ball leaves the barrel if you can ignore friction.

b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 5.20 N acts on the ball as it moves along the barrel.

c) For the situation in part B, at what position along the barrel does the ball have the greatest speed? (In this case, the maximum speed does not occur at the end of the barrel.)

d) What is that greatest speed?

Explanation / Answer

Given,

Spring constant k = 450 N/m

Spring is compressed to x = 6.4 cm or 0.064 m

mass of the ball = 0.027 kg

initially only stored energy is due to spring

finally only energy in ball is kinetic energy

by conservstion of energy

initial energy = final energy

0.5 * k * x^2 = 0.5 * m * v^2

450 * 0.064^2 = 0.027 * v^2

v = 8.26 m/s

a) the speed with which the ball leaves the barrel = 8.26 m/s

force due to spring = k * x

force = 450 * 0. 064

force = 28.8 N

the net force on the bullet is reduced by 5.20 N all along the barrel.

The force starts out at 28.8 and ends up at -5.2, for an average of 11.8.

11.8/14.4 * 68.27 = v^2

v = 7.47 m/s

b) speed of the ball as it leaves the barrel if a constant resisting force of 5.20 N acts on the ball as it moves along the barrel = 7.47 m/s

the velocity will be greatest where the net force acting will be 0 that is

450 * x = 5.20

x = 0.012

c) at position 0.012 m along the barrel does the ball have the greatest speed

average force = 14.4

so work done = 14.4 * 0.012

work done = 0.1728

which is equal to the kinetic energy

0.1728 = 0.5 * 0.027 * v^2

v = 3.58 m/s

d) that greatest speed = 3.58 m/s

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