Asked by Francis
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]
Answers
Answered by
Damon
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
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
Answered by
drwls
(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)
(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)
Answered by
drwls
Are you sure the pull back is only X = 0.05 meters?
Answered by
Francis
Thanks so much! Yeah, it's weird it says 0.05 meters. Thanks though!
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