A light spring has unstressed length of 17.1 cm. It is described by Hooke\'s law
ID: 1458004 • Letter: A
Question
A light spring has unstressed length of 17.1 cm. It is described by Hooke's law with spring constant 4.43 N/m. One end of the horizontal spring is held on a fixed vertical axle, and the other end is attached to a puck of mass mthat can move without friction over a horizontal surface. The puck is set into motion in a circle with a period of 1.40 s.
(a) Find the extension of the spring x as it depends on m.
x =
Evaluate x for the following masses. (If not possible, enter IMPOSSIBLE.)
(b) m = 0.0700 kg
m
(c) m = 0.140 kg
m
(d) m = 0.180 kg
m
(e) m = 0.220 kg
m
(f) Describe the pattern of variation of x as it depends on m.
Explanation / Answer
Given:-
The total distance is r = d + x
The unstretched length of the spring (d) = 0171m
Spring constat (k) = 4.43N/m
The time period isT = 1.40s
(a) The extension of the spring id given by
F = kx & ma = kx
m(v2/r) = kx
m((2r/T)2/r) = kx
m((42r2/T2)/r) = kx
(4m2r)/T2 = kx
(4m2d)/T2 + (4m2x)/T2 = kx
x = ((4m2d)/T2)/(k - (4m2)/T2)
x = ((4*m*2*0.171)/(1.4)2)/(4.43 - (4*m*2)/(1.4)2)
x = (4.29m)/(4.43 - 8.97m)
(b)
And for the mass (m) =0.070kg
Then the extension in the spring is
x = (4.29m)/(4.43 - 8.97m)
x = (4.29(0.07))/(4.43 - 8.97(0.07))
x = 0.0785m
(c)
And for the mass (m) =0.140kg
Then the extension in the spring is
x = (4.29m)/(4.43 - 8.97m)
x = (4.29(0.14))/(4.43 - 8.97(0.14))
x = 0.1872m
(d)
And for the mass (m) =0.180kg
Then the extension in the spring is
x = (4.29m)/(4.43 - 8.97m)
x = (4.29(0.18))/(4.43 - 8.97(0.18))
x = 0.2742m
(e)
And for the mass (m) = 0.190kg
Then the extension in the spring is
x = (4.29m)/(4.43 - 8.97m)
x = (4.29(.19))/(4.43 - 8.97(.19))
x = 0.300m
(f)
The extension is directly proportional to m when m is only a few grams. Then it grows faster and faster, diverging to infinity.
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