A light spring has unstressed length of 16.4 cm. It is described by Hooke\'s law

Question

A light spring has unstressed length of 16.4 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 m that can move without friction over a horizontal surface. The puck is set into motion in a circle with a period of 1.37 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

(c) m = 0.140 kg


(d) m = 0.180 kg
  

Explanation / Answer

Here,

k = 4.43 N/m
t = 1.37 s
l = 16.4 cm = 0.164 m

Spring force = centripital force

mw^2R = Kx
m(2*pi/T)^2 * (L + X) = Kx

m = K*x*T^2/(4*pi^2(L+x)) ----------------(1)

B)
m = 0.07 kg
Eqn (1) becomes

0.07 = 4.43*x*1.37^2/(4*3.14^2(0.164+x))

0.07 * (4*3.14^2(0.164+x)) = 4.43*x*1.37^2

2.76069 (x+0.164) = 8.31467 x

x = 0.08 m

C)
m = 0.140 kg
Eqn (1) becomes

0.140 = 4.43*x*1.37^2/(4*3.14^2(0.164+x))

0.140 * (4*3.14^2(0.164+x)) = 4.43*x*1.37^2

x = 0.32 m

D)
m = 0.180 kg
Eqn (1) becomes

0.180 = 4.43*x*1.37^2/(4*3.14^2(0.164+x))

0.180 * (4*3.14^2(0.164+x)) = 4.43*x*1.37^2

x = 0.95 m

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