A 4.07 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 144 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.5 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

To find the amplitude of the resulting simple harmonic motion, we can use the principle of conservation of mechanical energy.

The mechanical energy in the system is conserved because there are no external forces acting on the block-spring system (no friction, air resistance, or other forces). Therefore, the initial mechanical energy of the system will be equal to the final mechanical energy.

The initial mechanical energy of the system includes the kinetic energy of the block as it is shoved and the potential energy stored in the compressed spring. The final mechanical energy of the system is the potential energy stored in the stretched spring.

The initial kinetic energy (KE) of the block is given by the equation KE = 1/2 m v^2, where m is the mass of the block and v is the initial velocity.

KE = 1/2 * (4.07 x 10^-2 kg) * (10.5 m/s)^2
KE = 0.022224525 J

The potential energy (PE) stored in the compressed spring is given by the equation PE = 1/2 k x^2, where k is the spring constant and x is the displacement of the block from its equilibrium position (which is the amplitude in this case).

PE = 1/2 * (144 N/m) * x^2

Since the mechanical energy is conserved, the initial mechanical energy is equal to the final mechanical energy. Therefore:

0.022224525 J = 1/2 * (144 N/m) * x^2

Simplifying the equation:

0.022224525 J = 72 N/m * x^2

x^2 = 0.022224525 J / (72 N/m)

x^2 = 0.000308 Ave. m^2

Taking the square root of both sides to find the amplitude:

x = sqrt(0.000308 Ave. m^2)
x = 0.0175 m

Therefore, the amplitude of the resulting simple harmonic motion is 0.0175 meters.