If the system is frictionless there is no loss of energy, BUT the masses will definitely accelerate slower. The reason is twofold:
First, the component of the force of gravity that is pulling on the 4pound ball in the the direction of the ramp is smaller than if the 4pound weight is allowed to fall straight down. In fact, at some angle that force will precisely equal the force of gravity acting on the 3pound weight, and the system will no accelerate at all. And at lower ramp angles the system will run backward  the 3pound weight will cause the 4pound ball to roll uphill. The "magic angle" for this is arcsin(m/M) = 48.6 degrees. This comes from $\displaystyle \sum F = ma$:
$\displaystyle Mg \sin \theta  mg = (M+m) a $
Here M is the mass of the large ball, m is the mass of the smaller weight, and 'a' is the acceleration of the system. Positive values of 'a' means the ball is rolling down the ramp and the 3 pound weight is rising. If the left hand side = 0 then 'a' = 0 and the system is in equilibrium, and if the left hand side is negative then 'a' is negative.
Second, in order for the 4pound ball to roll (as opposed to slide down the ramp), some of the potential energy that is converted to kinetic energy is used to spin the ball (i.e it goes into rotational KE) and some is used to get the masses to moving linearaly:
$\displaystyle \Delta KE + \Delta PE = 0$
$\displaystyle Mg h \sin \theta  mgh = \frac 1 2 Mv^2 + \frac 1 2 m v^2 + \frac 1 2 I \omega^2$
Given that $\displaystyle \omega = v/R$:
$\displaystyle (M \sin \theta)gh = \frac 1 2 v^2(M+m+\frac I {R^2})$
Here 'h' is the vertical displacement upward of the 3pound weight. From this you can calculate velocity as a function of displacement of the weights. This can be simplified given that for a rolling ball $\displaystyle \omega = v/R$, and for a solid sphere $\displaystyle I = \frac 2 5 MR^2$:
$\displaystyle (M\sin \theta + m)gh = \frac 1 2 v^2(\frac {7M} 5 + m)$
