Question

A 3.49 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 111 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.3 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Answer 1

18.2 cm

Step-by-step explanation:

The kinetic energy of the block

= .5 x 3.49 x 10⁻² x 10.3²

= 1.85 J

If x be the compression created in the spring due to this kinetic energy

1.85 = 1/2 k x²

= .5 x 111 x²

x² = 3.3 x 10⁻²

x = 0.182 m .

= 18.2 cm

This will be the amplitude of oscillation under SHM.