A light spring has unstressed length of 15.1 cm. It is described by Hooke\'s law

Question

A light spring has unstressed length of 15.1 cm. It is described by Hooke's law with spring constant 4.5 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 m that can move without friction over a horizontal surface. The puck is set into motion in a circle with a period of 1.4 s.

A light spring has unstressed length of 15.1 cm. It is described by Hooke's law with spring constant 4.5 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 m that can move without friction over a horizontal surface. The puck is set into motion in a circle with a period of 1.4 s. (a) Find the extension of the spring x as it depends on m. x = Your response differs from the correct answer by more than 10%. Double check your calculations. m (e) m = 0.230 kg Your response differs from the correct answer by more than 10%. Double check your calculations. m (d) m = 0.180 kg Your response differs from the correct answer by more than 10%. Double check your calculations. m (c) m = 0.140 kg Evaluate x for the following masses. (If not possible, enter IMPOSSIBLE.) (b) m = 0.0700 kg

Explanation / Answer

Given that
The total distance is r = d + x
The unstretched length of the spring (d) = .155 m
Spring constat (k) = 4.3N/m
The time period isT = 1.3s
(a)
The extension of the spring id given by
F = kx
And
ma = kx
m(v2/r) = kx
m((2?r/T)2/r) = kx
m((4?2r2/T2)/r) = kx
(4m?2r)/T2 = kx
(4m?2d)/T2 + (4m?2x)/T2 = kx
x = ((4m?2d)/T2)/(k - (4m?2)/T2)
x = ((4*m*?2*.155)/(1.3)2)/(4.3 - (4*m*?2)/(1.3)2)
x = (3.62m)/(4.3 - 23.36m)

(b)
And for the mass (m) =0.070kg
Then the extension in the spring is
x = (3.62m)/(4.3 - 23.36m)
x = (3.62(.07))/(4.3 - 23.36(.07))
x = .0951m

(c)
And for the mass (m) =0.140kg
Then the extension in the spring is
x = (3.62m)/(4.3 - 23.36m)
x = (3.62(.14))/(4.3 - 23.36(.14))
x = .492m

(d)
And for the mass (m) =0.180kg
Then the extension in the spring is
x = (3.62m)/(4.3 - 23.36m)
x = (3.62(.18))/(4.3 - 23.36(.18))
x = 6.85m

(e)
And for the mass (m) = 0.190kg
Then the extension in the spring is
x = (3.62m)/(4.3 - 23.36m)
x = (3.62(.19))/(4.3 - 23.36(.19))
x = -4.97m
This situation is not possible.

(f)
The extension is directly proportional to m when m is only a few grams. Then it grows faster and faster, diverging to infinity for m

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