Answer
a) less than
b) $7.65m/s$
Work Step by Step
(a) We know that the sled starts with an initial speed of $1.5m/s$. Its final speed will be less than $9.00m/s$ because at the top of the hill the kinetic energy is small as compared to the gain in kinetic energy while the sled falls down the hill.
(b) We can find the required speed as follows:
$\frac{1}{2}mv^2_{bottom,2}+0=\frac{1}{2}mv^2_{top,2}+mgy_{top}=\frac{1}{2}mv^2_{top}+\frac{1}{2}mv^2_{bottom,1}$
This simplifies to:
$v_{bottom,2}=\sqrt{v_{top}^2+v^2_{bottom,1}}$
We plug in the known values to obtain:
$v_{bottom,2}=\sqrt{(1.50)^2+(7.50)^2}$
$v_{bottom,2}=7.65m/s$
