Problem 10.12

10.12.

First, let's find the velocity at point P.
From the conservation of energy,
K_A+U_A = K_P+U_P
$mgh = \frac{1}{2}mv_{P}^{2} + mgR$
$(6)(5)(9.8) J = \frac{1}{2}(6)(v_{P}^{2}) + (6)(9.8)(2)$
Solving for v_P²,
v_P = 58.8 m²/s²
Now, we are trying to find the components of the acceleration of the block at point P, the only forces acting on the system are the Normal force exerted by the track and the force of gravity. In the $\hat{y}$ direction,
a_y = a_t  =  g  =  9.8 m/s²
and in the $\hat{x}$ direction,
$a_{x} = a_{r} ~=~ \frac{v_{P}^{2}}{R} ~=~ 29.4 m/s^{2}$
where the subscripts r and t represent the radial and tangential components respectively.