A new villain is in town: The Brutal Flying Insect. Fortunately for the MArvel H

Question

A new villain is in town: The Brutal Flying Insect. Fortunately for the MArvel Heroes, a new superhero joined as well: the Fantastic Feynman: a brilliant scientist. Spidey has tried numerous times, although unsuccessfully to catch the Brutal Flying Insect. The super-brilliant mind of Feynman's has developed a plan based on his knowledge regarding charge acquired by flying insects, as the air friction strips electrons from them, and how natural spiderweb is using the Coulomb force to attract the insect.

In the Fantastic Feynmans advise, Spidey spins a large web ( in the vertical plane) between two tall buildings separated by 50.0 m that produce an electric field of 1.50 x 107 N/C perpendicular to the web. Shortly after, the brutal flying insect enters the "web zone" horizontally with a charge pf 50.0 x 10-6 C and speed of 30.0 m/s parallel to the web and perpendicular to the electric field. As the villain enters the zone, he is 5.00 m from the web. Does the web catch the villain, and if so, how deep into the "web zone" does the villain travel? Assume the mass of the Brutal Flying Insect is 68.5 kg and that the web does not move.

(Frictional Force due to web not given)

Explanation / Answer

horizontal speed = 30 m/s

there is no force in horizontal hence horiozontal speed will be constant .

and it have to travel 50 m horizontally.

time taken to reach the other building, t = 50 / 30 = 1.67 sec

In vertical, there are two force : electric and gravitational. (both will be vertically downwards)

Fnet = Fe + Fg = q E + m g

Fnet = (50 x 10^-6 x 1.50 x 10^7) + (68.5 x 9.8)

= 1421.3 N

a = Fnet / m = 20.75 m/s^2

vertical displacement in time t calculated above,

y = u t + at^2 /2 = 0 + 20.75 (1.667^2 ) /2 = 28.8 m

it was only 5 m above.

hence it will get caught.

how deep, d = 28.8 - 5 =23.8 m

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