A solid sphere rolls up an inclined plane of inclination angle 35 degrees. At the bottom of the incline, the center of mass of the sphere has a translational speed of 13 ft/s. How far (in feet) does the sphere roll up the plane (distance along the plane)?
To find the distance along the plane that the sphere rolls up, we need to consider the conservation of mechanical energy.
The total mechanical energy of the system is conserved, which is the sum of the kinetic energy (KE) and potential energy (PE). The KE of the sphere is due to its translational motion, and the PE is due to its position above the ground.
We can write the conservation of energy equation as follows:
Initial KE + Initial PE = Final KE + Final PE
Since the sphere is initially at the bottom of the inclined plane, the initial PE is zero. The initial KE is given by:
Initial KE = (1/2) * mass * velocity^2
where the mass is the mass of the sphere and the velocity is the translational speed at the bottom of the incline.
The final PE is given by:
Final PE = mass * gravitational acceleration * height
where the height is the distance along the inclined plane.
The final KE is zero because the sphere comes to rest at the top of the incline.
Combining these equations, we have:
(1/2) * mass * velocity^2 = mass * gravitational acceleration * height
Simplifying the equation, we find:
velocity^2 = 2 * gravitational acceleration * height
We can solve this equation for the height:
height = velocity^2 / (2 * gravitational acceleration)
Now we can substitute the given values:
velocity = 13 ft/s
gravitational acceleration = 32.2 ft/s^2 (approximation for acceleration due to gravity on Earth)
Substituting these values into the equation, we get:
height = (13 ft/s)^2 / (2 * 32.2 ft/s^2)
Evaluating the expression, we find:
height ≈ 3.10 ft
Therefore, the solid sphere rolls up the inclined plane a distance of approximately 3.10 feet.