Posted by Francis on Thursday, December 27, 2012 at 8:19pm.
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
(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)
Are you sure the pull back is only X = 0.05 meters?
Thanks so much! Yeah, it's weird it says 0.05 meters. Thanks though!
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