Question

A spring of force constant k = 3000 N/m launches 1-kg marbles upward

into a 20-kg block, as shown at right. The spring is compressed 0.2 m from
its natural length and then released. The marble travels upward 0.2 m as
the spring pushes it and then travels upward an additional 0.8 m to strike
the bottom of the block. The block then flies upward into the air.
106) What is the speed of the marble when it strikes the bottom of the
block?

Answer 1

The speed of the marble can be found using the equation for kinetic energy, KE = (1/2)mv^2, where m is the mass of the marble and v is its velocity. Since the marble is being launched by the spring, we can use the equation for potential energy, PE = (1/2)kx^2, to find its velocity.

The potential energy of the spring is (1/2)kx^2 = (1/2)(3000 N/m)(0.2 m)^2 = 600 J

The kinetic energy of the marble is KE = (1/2)mv^2, where m = 1kg, so KE = (1/2)(1 kg)(v^2).

So 600 J = (1/2)(1 kg)(v^2)

Solving for v, we get:

v = sqrt(1200 J/ 1kg)

v = sqrt(1200 m^2/s^2/ 1kg)

v = 34.64 m/s.

Step-by-step explanation:

Explained above