A smooth sphere of radius R is forced to translate along a straight...

Let the sphere accelerate horizontally with a constant acceleration $a$. We analyze the motion of the particle in the non-inertial frame of the sphere.

In this frame, the particle experiences a pseudo force $ma$ in the direction opposite to the sphere's acceleration, in addition to the downward gravitational force $mg$.

As the particle slides through an angle $heta$ from the top of the sphere, its displacement components relative to the sphere are:
Horizontal displacement in the direction of the pseudo force: $x = R\sin\theta$
Vertical displacement downwards: $y = R - R\cos\theta = R(1 - \cos\theta)$

Using the work-energy theorem in the frame of the sphere:
$W_{\text{net}} = \Delta K$

The normal reaction is always perpendicular to the path relative to the sphere, so it does zero work.
Work done by the pseudo force: $W_p = max = maR\sin\theta$
Work done by gravity: $W_g = mgy = mgR(1 - \cos\theta)$

The change in kinetic energy of the particle relative to the sphere is:
$\Delta K = \dfrac{1}{2}mv^2 - 0$

Equating the total work done to the change in kinetic energy:
$\dfrac{1}{2}mv^2 = maR\sin\theta + mgR(1 - \cos\theta)$

$\dfrac{1}{2}mv^2 = mR(a\sin\theta + g - g\cos\theta)$

$v^2 = 2R(a\sin\theta + g - g\cos\theta)$

$v = [2R(a\sin\theta + g - g\cos\theta)]^{1/2}$