A 620 nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of 0.14 eV are emitted. (a) Determine the work function (in eV). eV (b) Determine the cutoff frequency of the surface (in THz). THz (c) What is the stopping potential (in V) when the surface is illuminated with light of wavelength 420 nm

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j