# Question #4444b

I found 224Joules

#### Explanation:

Kinetic energy is equal to:

$K = \left(m \cdot {v}^{2}\right) \cdot \frac{7}{10}$

Assuming there are no losses, the speed is constant = 8m/s

So, $K = \left(7 \cdot {8}^{2}\right) \cdot \frac{7}{10}$ = 313,6 Joules

Oct 14, 2015

$313 , 6 J$

#### Explanation:

If we also take into account the rotational kinetic energy due to the rolling motion, then we get :

${\left({E}_{K}\right)}_{T o t a l} = {\left({E}_{k}\right)}_{T r a n s l a t i o n} + {\left({E}_{k}\right)}_{R o t a t i o n}$

$= \frac{1}{2} m {v}^{2} + \frac{1}{2} I {\omega}^{2}$, where I is the moment of inertia of the sphere, and omega is the angular velocity at which the sphere is rotating.

$= \left(\frac{1}{2} \times 7 \times {8}^{2}\right) + \left(\frac{1}{2} \times \frac{2}{5} \times 7 \times {R}^{2} \times {\omega}^{2}\right)$

$= \left(224 + \frac{7}{5} {R}^{2} {\omega}^{2}\right)$ Joules

where R is the radius of the solid sphere.

But since the linear and angular velocities may be related by the equation $v = r \omega$, we may substitute this into the expression to yield :

$= \left(224 + \frac{7}{5} {R}^{2} {\left(\frac{v}{R}\right)}^{2}\right) = 224 + \frac{7 \times {8}^{2}}{5} = 313 , 6 J$

Oct 17, 2015

KE =$\frac{1}{2} m {v}^{2}$