Answer
(a) $\theta = 12.3^{\circ}$
(b) $v = 1.59~m/s$
Work Step by Step
(a) We can find the block's acceleration.
$a = \frac{v^2-v_0^2}{2x} = \frac{(2.50~m/s)^2-0}{(2)(1.50~m)}$
$a = 2.083~m/s^2$
We can use a force equation to find the angle $\theta$.
$mg~sin(\theta) = ma$
$\theta = arcsin(\frac{a}{g}) = arcsin(\frac{2.083~m/s^2}{9.80~m/s^2})$
$\theta = 12.3^{\circ}$
(b) We can use a force equation to find the acceleration when there is friction.
$\sum F = ma$
$mg~sin(\theta) - 10.0~N = ma$
$a = \frac{(8.00~kg)(9.80~m/s^2)~sin(12.3^{\circ})- 10.0~N}{8.00~kg}$
$a = 0.8377~m/s^2$
We can find the speed $v$ at the bottom of the ramp.
$v^2 = v_0^2+2ax = 0+2ax$
$v = \sqrt{(2)(0.8377~m/s^2)(1.50~m)}$
$v = 1.59~m/s$
