Answer
(a) $a = 2.5~m/s^2$
(b) v = 6.3 m/s
Work Step by Step
(a) We can use a force equation to find the acceleration:
$ma = \sum F$
$ma = mg ~sin(\theta) - mg ~cos(\theta)\cdot ~\mu_k$
$a = g ~sin(\theta) - g ~cos(\theta)\cdot ~\mu_k$
$a = (9.80 ~m/s^2) ~sin(25.0^{\circ}) - (9.80 ~m/s^2) ~cos(25.0^{\circ})\cdot ~(0.19))$
$a = 2.45~m/s^2$
(b) We can use the acceleration to find the speed at the bottom of the inclined plane:
$v^2 = v_0^2 + 2ax = 0 + 2ax = 2ax$
$v = \sqrt{2ax} = \sqrt{(2)(2.45 ~m/s^2)(8.15 ~m)}$
$v = 6.32 ~m/s$
