Asked by Emily
If the spring is compressed a distance of 0.150 m and the gun fired vertically as shown, the gun can launch a 27.6g projectile from rest to a maximum height of 16.1 m above the starting point of the projectile. Neglecting all resistive forces, determine the spring constant.
Determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0) (as shown in (b)).
Determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0) (as shown in (b)).
Answers
Answered by
bobpursley
You know the energy in the spring:
PE=massprojetile*g*heightitgoes
so now in the spring.
PE=1/2 k x^2
solve for k
speed through equilibrium? 1/2 mv^2=PE above solve for v
PE=massprojetile*g*heightitgoes
so now in the spring.
PE=1/2 k x^2
solve for k
speed through equilibrium? 1/2 mv^2=PE above solve for v
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