Posted by **Francis** on Thursday, December 27, 2012 at 8:19pm.

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]

- Physics -
**Damon**, Thursday, December 27, 2012 at 8:39pm
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

- Physics -
**drwls**, Thursday, December 27, 2012 at 8:46pm
(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)

- Physics -
**drwls**, Thursday, December 27, 2012 at 8:48pm
Are you sure the pull back is only X = 0.05 meters?

- Physics -
**Francis**, Thursday, December 27, 2012 at 9:15pm
Thanks so much! Yeah, it's weird it says 0.05 meters. Thanks though!

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