Question

4. A spring with k=45 N/m is oriented vertically with one end fixed to the ground. A 0.5 kg mass on top of the spring compresses it. 
 Provide a conceptual explanation of what is happing for each of the following cases a. You hold the mass while you gently compress the spring, and when you release the mass it sits at rest atop the spring. b. You place the mass on the uncompressed spring and release it. c. You drop the mass from 10 cm above the spring.

Answer 1

Case a )

If we gently keep mass over spring then spring will compress , due to weight of mass,

Let spring compress by distance x, so Restoring force by spring =- kx ,and it is balanced by mg so mass is at rest on top of spring.

so mg = -kx =====> 0.5 * 9.8 = 45 * x ====> So x= 0.1088 m = 10.88 cm

SO Spring will compress by   10.88 cm .

Case b)

If we will realease mass after placing so it will apply weight =4.9 N , SO spring will compressed by 10.88 cm

and Potential energy will by stored inside the spring due to compression of 1/2 Kx² = 1/2 *45 * (.1088)²

So P.E = .266 J , SO Object will start osicallation in SHM form but energy will be lossed and it will be damped after sometime.

Case 3)

If we drop from hieght 10 cm , So Energy of mass = mgh = 0.5 * 9.8 * 0.1 m = 0.49 J

So this Energy will be transferred to Spring , So 1/2 K x^{2 =}    0.49 J ===> So x= .1475 m = 14.75 cm

SO Spring will displaced by 14.75 cm , and it will start oscillating.