Answer
(a) $42Kg$
(b) $0.74m$
Work Step by Step
(a) We can find the student's mass as follows:
$F_1+F_2=mg$
This can be rearranged as:
$m=\frac{F_1+F_2}{g}$
We plug in the known values to obtain:
$m=\frac{290N+122N}{9.81m/s^2}$
$m=42Kg$
(b) We can find the required distance as follows:
$x=\frac{F_2 L}{mg}$
We plug in the known values to obtain:
$x=\frac{(122N)(2.50m)}{(42Kg)(9.81m/s^2)}$
$x=0.74m$
