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s34n6962

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The homework template should be left in the post.

A 0.150-kg
frame, when suspended from a coil spring, stretches the spring 0.070 m
. A 0.200-kg
lump of putty is dropped from rest onto the frame from a height of 30.0 cm

Find the maximum distance the frame moves downward from its initial position.

To start off the question, I found the Spring Constant of the spring using the fact that if the spring were at equilibrium .070 m away from its relaxed point, the forces of gravity (F=mg) and the force of the spring (F=-kx), should sum up to 0. Therefore mg=kx; so (.15 kg)(9.8 m/s^2)=k(.07 m) so k=21.

Then I found the energy held between the putty and the Earth in potential energy U=mgy; U=(.2 kg)(9.8 m/s^2)(.3 m); and used that to find what it's velocity would be just before it strikes the frame so U=K; (.2 kg)(9.8 m/s^2)(.3 m)=(1/2)(.2 kg)(v)^2; v=((2)(.2)(9.8))^1/2 or roughly 2.42 m/s.

From there I used momentum conservation since the interaction between the putty and frame have outside frictional forces that do work, destroying energy conservation, but retains momentum conservation since the forces are balanced. So mv=mv; (.2 kg)(2.42 m/s)=(.35 kg)v; roughly v=1.39 m/s

Now that I know how fast the interaction between the objects will cause the objects to go I can reapply energy conservation since there are no longer any outside forces that do work on our system, which consists of the frame, the spring, the putty, and earth. So we should obtain the equation (1/2)(.35 kg)(1.39 m/s)+(1/2)(21)(.07 m)^2=(1/2)(21)(x-.07 m)^2-(.35 kg)(9.8 m/s^2)(x), the first term being the kinetic energy of the moving frame/putty, the second being the energy initially stored in the spring, the third being the energy gained by the spring by the interaction, and the fourth being the potential energy lost with gravity since it is moving down. This equation can ultimately be simplified as 10.5x^2-1.96x-.38255=0, which yields a value of x=3.058055537 m, which MasteringPhysics says is very close but that I have a rounding issue, it would be much appreciated if someone would help me mend this miniscule error as it does not allow for partial credit to be given, and it should be noted that I didn't use the rough values to calculate this but rather put in as many as my calculator would give me so I don't believe that that's where my error comes from.

Find the maximum distance the frame moves downward from its initial position.

To start off the question, I found the Spring Constant of the spring using the fact that if the spring were at equilibrium .070 m away from its relaxed point, the forces of gravity (F=mg) and the force of the spring (F=-kx), should sum up to 0. Therefore mg=kx; so (.15 kg)(9.8 m/s^2)=k(.07 m) so k=21.

Then I found the energy held between the putty and the Earth in potential energy U=mgy; U=(.2 kg)(9.8 m/s^2)(.3 m); and used that to find what it's velocity would be just before it strikes the frame so U=K; (.2 kg)(9.8 m/s^2)(.3 m)=(1/2)(.2 kg)(v)^2; v=((2)(.2)(9.8))^1/2 or roughly 2.42 m/s.

From there I used momentum conservation since the interaction between the putty and frame have outside frictional forces that do work, destroying energy conservation, but retains momentum conservation since the forces are balanced. So mv=mv; (.2 kg)(2.42 m/s)=(.35 kg)v; roughly v=1.39 m/s

Now that I know how fast the interaction between the objects will cause the objects to go I can reapply energy conservation since there are no longer any outside forces that do work on our system, which consists of the frame, the spring, the putty, and earth. So we should obtain the equation (1/2)(.35 kg)(1.39 m/s)+(1/2)(21)(.07 m)^2=(1/2)(21)(x-.07 m)^2-(.35 kg)(9.8 m/s^2)(x), the first term being the kinetic energy of the moving frame/putty, the second being the energy initially stored in the spring, the third being the energy gained by the spring by the interaction, and the fourth being the potential energy lost with gravity since it is moving down. This equation can ultimately be simplified as 10.5x^2-1.96x-.38255=0, which yields a value of x=3.058055537 m, which MasteringPhysics says is very close but that I have a rounding issue, it would be much appreciated if someone would help me mend this miniscule error as it does not allow for partial credit to be given, and it should be noted that I didn't use the rough values to calculate this but rather put in as many as my calculator would give me so I don't believe that that's where my error comes from.