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A 0.359-kg box is placed in contact with a spring of stiffness 1.343×103 N/m. Th

ID: 1462979 • Letter: A

Question

A 0.359-kg box is placed in contact with a spring of stiffness 1.343×103 N/m. The spring is compressed 6.00×10-2 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 1.360 m above the floor.

(the drawing is not to scale) What is the potential energy stored in the spring when it is compressed?
2.42 J

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?
2.42 J

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

How much time elapses between the time when the block leaves the table and the time just before the block hits the floor?
5.27×10-1 s

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

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Explanation / Answer

a)

the potential energy stored in the spring = 1/2*1.343*10^3*(6*10^-2)^2 = 2.41 J

b)

kinetic energy of the block just before it leaves the table = 2.41 J

c)

2.41 + 0.359*9.8*1.360 = kinetic energy of the block just before it hits the floor

= 7.194 J

d)

T = sqrt((2*1.360)/(9.8)) = 0.526 sec

e)

2.41 = 1/2*0.359*V^2

=> V = 3.66 m/s just it leaves spring

=> d = 0.526* 3.66 = 1.92 m