Answer
Thus, the graph for $h \geq 2.5R = 0.30 m$ consists of a straight line of positive slope $2mg/R$ (which can be set to some convenient values for graphing purposes). Note that for $h \leq 2.5R,$ the normal force is zero.
Work Step by Step
The normal force $F_{N}$ for speeds $v_{t}$ greater than $\sqrt(gR)$ (whic possibilities for nonzero $F_{N}-$see the solution in the previous part), obeys
$F_{N} = \frac{mv_{t}^{2}}{R} - mg$
from Newton's second law. Since $v_{t}^{2}$ is related to $h$ by energy conservation
$K_{p} + U_{p} = K_{t} + U_{t} =\gt gh = \frac{1}{2} v_{t}^{2} + 2gR$
then the normal force, as a function for $h$ (so long as $h \geq 2.5R -$see the solution in the previous part), becomes
$F_{N} = \frac{2mgh}{R} - 5mg$
Thus, the graph for $h \geq 2.5R = 0.30 m$ consists of a straight line of positive slope $2mg/R$ (which can be set to some convenient values for graphing purposes). Note that for $h \leq 2.5R,$ the normal force is zero.
