A 0.361-kg box is placed in contact with a spring of stiffness 2.818x103 N/m. Th

Question

A 0.361-kg box is placed in contact with a spring of stiffness 2.818x103 N/m. The spring is compressed 6.70x10-2 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 1.130 m above the floor. (the drawing is not to scale) What is the potential energy stored in the spring when it is compressed? Submit Answer Incorrect. Tries 1/10 Previous Tries 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? Submit Answer Tries 0/10 What is the kinetic energy of the block just before it hits the floor? Submit Answer Tries 0/10 How much time elapses between the time when the block leaves the table and the time just before the block hits the floor? Submit Answer Tries 0/10 What distance, d, from the edge of the table does the block hit the floor?

Explanation / Answer

here,

mass , m = 0.361 kg

spring constant , K = 2818 N/m

d = 0.067 m

h = 1.13 m

a)

the initial potential energy of the spring , PEi = 0.5 * k * d^2

PEi = 0.5 * 2818 * 0.067^2 J

PEi = 6.33 J

b)

the kinetic energy of block just it leaves the spring , KEi = initial potential energy

KEi = 6.33 J

c)

the final kinetic energy , KEf = KEi + m * g * h

KEf = 6.33 + 0.361 * 9.81 * 1.13 J

KEf = 10.33 J

d)

let the time taken to hit the floar be t

h = 0 + 0.5 * g * t^2

1.13 = 0 + 0.5 * 9.81 * t^2

solving for t

t = 0.48 s

e)

horizontal speed be vx

0.5 * m * vx^2 = KEi

0.5 * 0.361 * vx^2 = 6.33

vx = 5.92 m/s

d = vx * t

d = 5.92 * 0.48 = 2.84 m

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