Due this Friday, Nov 13 at 11:59 pm (EST) A 0.272-kg box is placed in contact wi

Question

Due this Friday, Nov 13 at 11:59 pm (EST)

A 0.272-kg box is placed in contact with a spring of stiffness 4.352×102 N/m. The spring is compressed 1.08×10-1 m from its unstrained length. The spring is then released, the block slides across a frictionless tabletop, and it flies through the air. The tabletop is a height of 1.060 m above the floor.
[mass_spring_table]
(the drawing is not to scale) What is the potential energy stored in the spring when it is compressed?

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What is the kinetic energy of the block just before it leaves the table but after it is no longer in contact with the spring?

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What is the kinetic energy of the block just before it hits the floor?

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How much time elapses between the time when the block leaves the table and the time just before the block hits the floor?

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What distance, d, from the edge of the table does the block hit the floor?

Explanation / Answer

POtential energy of spring = k x^2 /2

         = (4.352 x 10^2) (1.08 x 10^-1)^2 /2 = 2.54 J

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using energy conservation,

KE = 2.54 J

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using energy conservation again,

initial total energy = final total energy

2.54 + ( 0.272 x 9.81 x 1.060) = KE + 0

KE = 5.37 J

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KE at top of table = 2.54 = mv^2 /2

2 x 2.54 = 0.272 x v^2

v = 4.32 m/s

initial velocity was in horizontal.

In vertical:

no initial velocity is vertical direction.
it has to go 1.06 m downward.

using h = ut + at^2 /2

1.06 = 0 + 9.81t^2 /2

t = 0.465 sec

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d = vt = 4.232 x 0.465 = 2 m

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