A(hypothetical) large slingshot is stretched 4.00 m to launch a 440 g projectile with speed sufficient to escape from earth (11.2 km/s). assume the elastic slingshot obeys hooke's law. (a) what is the spring constant of the device, if all the elastic potential energy is converted to kinetic energy? incorrect: your answer is incorrect. n/m (b) assume that an average person can exert a force of 220 n. how many people would be required to stretch the slingshot? people

(A) 3,449,600 N/m

(B) 62,720 poeple

Explanation:

extension of the sling shot (e) = 4 m

mass of projectile (m) = 440 g = 0.44 kg

speed of projectile (v) = 11.2 km/s = 11,200 m/s

average force a person can exert = 220 N

spring constant (k) = ?

(A) When all the elastic potential energy is converted to kinetic energy

rearranging the above equation

spring constant (K) =

spring constant (K) =

spring constant (K) =

spring constant (K) = 3,449,600 N/m

(B) force required to stretch the slingshot (F) = ke = 3,449,600 x 4 = 13,798,400 N

number of people required = required force / average force per person = 13,798,400 / 220 =62,720 poeple

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.