A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm s$$^{-1}$$. Its kinetic energy is:

We are given a uniform solid sphere having mass $$m = 500 \text{ g} = 0.5\ \text{kg}$$. It rolls without slipping on a horizontal surface, and the linear speed of its centre is $$v = 5.00\ \text{cm s}^{-1} = 0.05\ \text{m s}^{-1}$$.

For a rolling body, the total kinetic energy is the sum of translational and rotational parts. We therefore write the basic relation

$$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}}.$$

According to the definition, the translational kinetic energy is

$$K_{\text{trans}} = \dfrac12 m v^{2}.$$

For rotational kinetic energy about the centre of mass, the standard formula is

$$K_{\text{rot}} = \dfrac12 I \omega^{2},$$

where $$I$$ is the moment of inertia of the body about its centre and $$\omega$$ is the angular speed.

Because the sphere is uniform (solid), its moment of inertia about its centre is well-known:

$$I = \dfrac{2}{5} m r^{2}.$$

Rolling without slipping provides the kinematic link

$$\omega = \dfrac{v}{r},$$

since each point on the rim that touches the ground momentarily has zero velocity relative to the ground.

Now we substitute $$I$$ and $$\omega$$ into the rotational energy expression:

$$K_{\text{rot}} = \dfrac12 \left(\dfrac{2}{5} m r^{2}\right) \left(\dfrac{v}{r}\right)^{2} = \dfrac12 \left(\dfrac{2}{5} m r^{2}\right) \dfrac{v^{2}}{r^{2}} = \dfrac12 \left(\dfrac{2}{5}\right) m v^{2} = \dfrac{1}{5} m v^{2}.$$

Hence the total kinetic energy becomes

$$K_{\text{total}} = \dfrac12 m v^{2} + \dfrac15 m v^{2} = \left(\dfrac12 + \dfrac15\right) m v^{2} = \left(\dfrac{5}{10} + \dfrac{2}{10}\right) m v^{2} = \dfrac{7}{10} m v^{2}.$$

Substituting the numerical values $$m = 0.5\ \text{kg}$$ and $$v = 0.05\ \text{m s}^{-1}$$, we get

$$K_{\text{total}} = \dfrac{7}{10}\,(0.5)\,(0.05)^{2}.$$

First we square the speed:

$$(0.05)^{2} = 0.0025.$$

Next we multiply step by step:

$$\dfrac{7}{10} \times 0.5 = 0.7 \times 0.5 = 0.35,$$

and then

$$0.35 \times 0.0025 = 0.000875.$$

Expressing this result in scientific notation, we write

$$K_{\text{total}} = 8.75 \times 10^{-4}\ \text{J}.$$

Examining the options, this numerical value matches Option A.

Hence, the correct answer is Option A.