A solid sphere of mass 4.0 kg and radius 0.12 m is at rest at the top of a ramp inclined 15 $\degree$.It rolls to the bottom without slipping.The upper end of the ramp is 1.2 m higher than the lower end.Find the sphere's total kinetic energy when it reaches the bottom.
A)70 J
B)47 J
C)18 J
D)8.8 J

Question

Quizplus · Accepted Answer

First, we need to find the speed at the bottom of the ramp. Using conservation of energy, we can find the potential energy at the top and convert it into kinetic energy at the bottom:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Where $m$ is the mass of the sphere, $h$ is the height difference between the top and bottom of the ramp, $v$ is the speed at the bottom of the ramp, $I$ is the moment of inertia of the sphere, and $\omega$ is the angular speed of the sphere.
Since the sphere rolls without slipping, we can use the relationship between linear and angular speed, $v = \omega r$, where $r$ is the radius of the sphere. We can also use the formula for the moment of inertia of a solid sphere, $I = \frac{2}{5}mr^2$. Substituting these equations into the conservation of energy equation, we get:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$
Simplifying this equation and solving for $v$, we get:
$v = \sqrt{\frac{10gh}{7}}$
Plugging in the given values, we get $v \approx 1.53$ m/s. 
The total kinetic energy of the sphere at the bottom is the sum of its translational and rotational kinetic energies:
$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Again using the relationship between linear and angular speed, we can write $\omega = \frac{v}{r}$. Plugging in the given values and solving, we get $K \approx 47$ J. Therefore, the best choice is B.

A Solid Sphere of Mass 4 °\degree° It Rolls to the Bottom Without Slipping

A Solid Sphere of Mass 4 $\degree$ It Rolls to the Bottom Without Slipping