Question

A spring-loaded gun has a spring for which

Answer 1

Answer:

7.5 m

Explanation:

k = Spring constant = 180 N/m

x = Displacement of spring = 14 cm

m = Mass of projectile = 0.024 kg

a = g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

s = Displacement of projectile

v = Final velocity = 0

u = Initial velocity

The potential energy of the spring will be equal to the kinetic energy of the object

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mu^2\\\Rightarrow u=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow u=\sqrt{\dfrac{180\times 0.14^2}{0.024}}\\\Rightarrow u=12.12\ \text{m/s}$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-12.12^2}{2\times -9.81}\\\Rightarrow s=7.5\ \text{m}$

The maximum height reached above the initial position is 7.5 m