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Question: A mass of 0.3 Kghangs motionless from a vertical spring whose length is 0.8 mand whose unstretched length is 0.65 m. Next the mass is pulled down so the spring has a length of 0.9 mand is given an initial speed upward of 1.2 m/s. What is the maximum length of the spring during the following motion? What approximations or simplifying assumptions did you make?

Short Answer

Expert verified

The maximum length obtained for the spring during the motion is lMAX=0.94m.

Step by step solution

01

Definition of the displacement

The term displacement refers to a shift in an object's location. It's a vector quantity with a magnitude and a direction. It's depicted as an arrow pointing from the beginning place to the destination.

The force in a spring is given by:

F=kx

Here kis spring constant and xis the distance from the equilibrium point.

02

Finding the value of spring constant

Letl1is the unstretched length of the spring andl2is the length of the vertical spring.

Substitutex=l2-l1into the formula of spring force.

F=kl2-l1

The downward force will be F=mg, where m is the mass, g is the acceleration due to gravity whose value is 9.8m/s2.

SubstituteF=mginto the obtained formula and solve for k.

mg=kl2-l1k=mgl2-l1

Substitute m = 0.3 kg, g=9.8m/s2, l2=0.8mand l1=0.65minto the obtained result.

k=0.3kg(9.8m/s2)0.8m-0.65m=19.6N/m

Therefore, the value of spring constant is 19.6 N/m.

03

Finding maximum displacement

Letvf is the final speed,vjis the initial speed,dmis the maximum displacement,lsis the stretched length andl1is the unstretched length.

Apply the conservation of energy to find the maximum lengthlm.

Ej=EfKEf+PEf=KEj+PEj12mvf2+12kdm2=12mvj2+12kls=l12

Substitute m = 0.3 kg, vf=0, k = 19.6 N/m, vi=1.2m/s2, g=9.8m/s2, ls=0.9mand l1=0.65minto the obtained result and solve for

0+12(19.6)(dm)2=12(0.3)(1.2)2+12(19.6)(0.9-0.65)212(19.6)(dm)2=0.216+0.6125(dm)2=0.82859.8dm=0.29m

Therefore, the maximum displacement is 0.29 m.

04

Finding maximum length

Add the value of the maximum displacement,dmand the unstretched lengthl1to get the maximum length.

lMAX=dm+l1=0.29m+0.65m=0.94m

Therefore, the maximum length is 0.94 m.

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