A ball (m=0.5 kng, I = 2/5MR^2) is held against a spring compressed by a distance x= 12 cm on the ground and released

spring-constant

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A ball (m=0.5 kng, I = 2/5MR^2) is held against a spring compressed by a distance x= 12 cm on the ground and released. The balls roll without slipping along the track and leaves the track moving straight up at H0 = 2.5m. THen the ball freefalls upward to a height H1= 3.8 m. Find the spring constant , k, of the spring

Answer:

The driving principles behind this problem are conservation of energy and kinematics.
The stored elastic potential energy from the spring is imparted onto the ball. That energy is converted into rolling and translational kinetic energy.
$E_{0}=E_{f}$

$\dfrac{1}{2}kx^{2}=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}Iw^{2}$

The ball leaves the ramp and goes straight upward and loses all of its velocity in 3.8-2.5 = 1.3 meters. Use kinematics to find the numerical value of the translational velocity.
Finally use that fact that the moment of inertia is given and that v=rw to simplify the conservation of energy expression. You’ll find that all of the values are given and you can easily solver for the spring constant.

k=1200 N/m