Answer:
k = 45.95 N/m
Explanation:
First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.
[tex]R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\[/tex]
where,
Vo = Launch Speed = ?
R = Horizontal Range = 5.3 m
θ = Launch Angle = 35°
Therefore,
[tex]v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\[/tex]
v₀² = 55.33 m²/s²
Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:
[tex]Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\[/tex]
where,
k = spring constant = ?
x = compression = 17 cm = 0.17 m
m = mass of ball = 24 g = 0.024 kg
Therefore,
[tex]k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\[/tex]
k = 45.95 N/m
