Question

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

Answer 1

Answer:

(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

$o.5ke^{2} = 0.5 mv^{2}$

rearranging the above equation

spring constant (K) = $\frac{0.5mv^{2} }{0.5e^{2}}$

spring constant (K) = $\frac{0.5x0.44x11,200^{2} }{0.5x4^{2}}$

spring constant (K) = $\frac{27,596,800}{8}$

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