A light spring has an unstressed length of 15.5 cm. It isdescribed by Hooke\'s l

Question

A light spring has an unstressed length of 15.5 cm. It isdescribed by Hooke's law with spring constant 4.30 N/m. Oneend of the horizontal spring is held on a fixed vertical axle, andthe other end is attached to a puck of mass m that can move withoutfriction over a horizontal surface. The puck is set intomotion in a circle with a period of 1.30 s. Find theextension of the spring x as it depends on m. Evaluate x for(b) m = 0.070 kg, (c) m = 0.140 kg, (d) m=0.180 kg and (e) m= 0.190kg.

I am stuck on where to begin this. I need to find theextension of the spring x as it depends on m. Usem as necessary. After that, I am sure I can do it



Thanks in advance for any help

Explanation / Answer

Let the extension be x m , Then radius of the circle that the puck makes =0.155+ x m Also if the extension is then from Hooke's law , Restoring force in spring =kx = 4.30x N Time period of motion = 1.30 sec so linear velocity (v)= 2r/T = 2(0.155+x) /1.30............(i) In this case the required centripetal force is provided by therestoring force of spring also centripetal force =mv2/r Putting all the values mv2/r = 4.30x Put v=2r/T this will make it m(42r2)/T2r = 4.30x =>4m2r/T2 = 4.30x Putting the values =>23.36mr = 4.30 x Putting r = 0.155+x =>23.36m(0.155+x) =4.30x This is a linear equation in x and m Put different values of m to find corresponding x .

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