Answer
(a) v = 16 m/s
(b) v = 13 m/s
Work Step by Step
If the slope is 1-in-4 then $\theta = sin^{-1}(\frac{1}{4})$. The angle of the road is $\theta = 14.5^{\circ}$
(a) We can use the force equation to find the acceleration.
$ma = mg ~sin(\theta)$
$a = g ~sin(\theta) = (9.80 ~m/s^2)~sin(14.5^{\circ}) = 2.45 ~m/s^2$
We can use the acceleration to find the speed $v$ at the bottom.
$v^2 = v_0^2 + 2ax = 0 + 2ax = 2ax$
$v = \sqrt{2ax} = \sqrt{(2)(2.45 ~m/s^2)(55 ~m)} = 16 ~m/s$
(b) We can use the force equation to find the acceleration.
$ma = mg ~sin(\theta) - mg ~cos(\theta) ~\mu_s$
$a = g ~sin(\theta) - g ~cos(\theta) ~\mu_s$
$a = (9.80 ~m/s^2)~sin(14.5^{\circ}) - (9.80 ~m/s^2)~cos(14.5^{\circ}) \cdot (0.10))$
$a= 1.5 ~m/s^2$
We can use the acceleration to find the speed $v$ at the bottom.
$v^2 = v_0^2 + 2ax = 0 + 2ax = 2ax$
$v = \sqrt{2ax} = \sqrt{(2)(1.5 ~m/s^2)(55 ~m)} = 13 ~m/s$