Question

A slingshot with k = 400. N/m is pulled back 0.0500 m to shoot a 0.100-kg stone. If the slingshot projects the stone straight up in the air, (a) what is the speed when the stone is first released, and (b) what is the maximum height to which the stone will rise? [neglect friction]

Answer 1

potential energy = (1/2)(400)(.5)^2 = 50 J

so Ke = 50 = (1/2) m v^2
(1/2)(.1)v^2 = 50
v = 31.6 m/s

potential energy at top = m g h = 50
.1 * 9.8 * h = 50
h = 51 meters

Answer 2

(a) (Stored slingshot potential energy) = (Kinetic energy of stone when released)

(1/2)kX^2 = (1/2)MVo^2

Vo = X*sqrt(k/M)
= 0.05*sqrt(400/0.1) = 3.16 m/s
(which is not very fast)

(b) (Maximum gravitational potential energy) = (Stored slingshot potential energy)
M g H = (1/2) k X^2
H = [k/(2Mg)]*X^2
(not very high, either)

Answer 3

Are you sure the pull back is only X = 0.05 meters?

Answer 4

Thanks so much! Yeah, it's weird it says 0.05 meters. Thanks though!