Answer
The normal force on the block is $~~44.7~N$
Work Step by Step
We can find the mass of the block:
$\frac{1}{2}mv^2 = 40.0~J$
$m = \frac{(2)(40.0~J)}{v^2}$
$m = \frac{(2)(40.0~J)}{(4.00~m/s)^2}$
$m = 5.0~kg$
We can use the magnitude of the work done by gravity to find the angle $\phi$ that the ramp makes with the vertical:
$W = mgd~cos~\phi = 40.0~J$
$cos~\phi = \frac{40.0~J}{mgd}$
$\phi = cos^{-1}(\frac{40.0~J}{mgd})$
$\phi = cos^{-1}[\frac{40.0~J}{(5.0~kg)(9.8~m/s^2)(2.0~m)}]$
$\phi = cos^{-1}(0.408)$
$\phi = 65.9^{\circ}$
We can find the normal force on the block:
$F_N = mg~sin~\phi$
$F_N = (5.0~kg)(9.8~m/s^2)~sin~65.9^{\circ}$
$F_N = 44.7~N$
The normal force on the block is $~~44.7~N$
