Question

A mass of m = 0.250 kg is attached to the end of a massless spring of unknown spring constant. The mass is dropped from rest at point A, with the spring initially unstretched at a length of l1 = 1.50 m. As the mass falls, the spring stretches to a length of l2 = 2.30 m. At point B the mass is as shown in the diagram. (a) Given that the velocity vf = 5.80 m/s, find the force constant of the spring. (b) Find the acceleration of the mass at point B. (Define the positive direction to be upward.)

          A mass of m = 0.250 kg is attached to the
end of a massless spring of unknown spring constant. The mass is
dropped from rest at point A, with the spring initially unstretched
at a length of l1 = 1.50 m. As the mass falls,
the spring stretches to a length
of l2 = 2.30 m. At point B the mass is as shown
in the diagram. (a) Given that the
velocity vf = 5.80 m/s, find the force constant
of the spring. (b) Find the acceleration of the mass at point
B. (Define the positive direction to be upward.)
        
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Added by Miriam E.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A mass of m = 0.250 kg is attached to the end of a massless spring of unknown spring constant. The mass is dropped from rest at point A, with the spring initially unstretched at a length of l1 = 1.50 m. As the mass falls, the spring stretches to a length of l2 = 2.30 m. At point B the mass is as shown in the diagram. (a) Given that the velocity vf = 5.80 m/s, find the force constant of the spring. (b) Find the acceleration of the mass at point B. (Define the positive direction to be upward.)
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Transcript

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00:01 All right, so let's say we have a spring that has kind of dropped and stretched as it rotates through these two angles.
00:08 And it starts off with an initial distance of one point, or initial length of 1 .5 meters and stretches to a length of 2 .3 meters.
00:20 And the mass on the end of the spring is a quarter of a kilogram.
00:25 And at the bottom it has a speed total of 5 .8 meters per second.
00:32 So we want to know what is the spring constant, first off.
00:36 So what we'll have is the potential energy of the spring, mgh, the initial potential energy, is going to equal the spring energy, 1fkx squared, plus the kinetic energy at the bottom of its path.
00:53 And so the height that it falls through is going to be like 2 .3 meters.
00:59 So first off, let's write 1 .5 k x squared is equal to.
01:04 And we can write it as m times gh minus 1 half v squared.
01:10 So k is going to be 2m over x squared times gh minus 1 half v squared...
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