A solid uniform sphere is released from the top of an inclined plane .25 m tall. the sphere rolls down the plane without slipping and there is no energy lost from friction. What is the translational speed of the sphere at the bottom of the incline?

Ke = (1/2) m v^2 + (1/2) I omega^2

v = omega r
so omega = v/r

Ke = (1/2) m v^2 + (1/2) I v^2/r^2

= m g (.25)

(1/2) m v^2 + (1/2) (2/5) m v^2 = .25 m g

(7/10) v^2 = .25 * 9.81

To find the translational speed of the sphere at the bottom of the incline, we can make use of the principles of conservation of energy.

First, let's define some variables:
- m: mass of the sphere
- h: height of the incline (0.25 m in this case)
- R: radius of the sphere
- v: translational speed of the sphere at the bottom of the incline
- g: acceleration due to gravity (9.8 m/s²)

Given that the sphere rolls down the plane without slipping, both translational and rotational kinetic energy are involved.

The total mechanical energy of the sphere at the top of the incline is equal to the total mechanical energy at the bottom. The mechanical energy can be expressed as the sum of translational and rotational kinetic energy:

At the top:
E_top = mgh

At the bottom:
E_bottom = 1/2 mv² + 1/2 Iω²

Since the sphere rolls down without slipping, the linear velocity v is related to the angular velocity ω by the equation v = Rω, where R is the radius of the sphere. Additionally, the moment of inertia I of a solid sphere is given by I = 2/5 mR².

Substituting the values into the equations:

mgh = 1/2 mv² + 1/2 (2/5 mR²) (v/R)²

Simplifying the equation:

gh = 1/2 v² + 1/5 v²
gh = 7/10 v²

Solving for v:

v² = (10/7)gh

Taking the square root:

v = √((10/7)gh)

Now we can substitute the known values:
g = 9.8 m/s²
h = 0.25 m
R (radius) = ?
m (mass) = ?

Since the question does not provide those values, we cannot determine the specific translational velocity. However, if we know the values of the radius and mass, we can use the equation v = √((10/7)gh) to calculate the translational speed of the sphere at the bottom of the incline.