Question

as a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change?
the velocity decreases,
the acceleration increases,
and the net force
increases.
the velocity decreases,
the acceleration
decreases, and the net
force decreases.
the velocity increases, the
acceleration decreases,
and the net force
decreases.
the velocity increases, the
acceleration increases,
and the net force
decreases.

Explanation:

For a mass - spring system, we use Hooke's Law \( F=-kx \) (where \( F \) is the restoring force, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium) and Newton's second law \( F = ma \) (where \( m \) is mass and \( a \) is acceleration). As the mass moves farther from equilibrium (\( x \) increases), the magnitude of the restoring force \( F \) (net force here) increases because \( F\propto|x| \). From \( F = ma \), since \( m \) is constant, acceleration \( a\propto F \), so acceleration increases. Regarding velocity, when the mass moves away from equilibrium, it is moving against the restoring force, so the force does negative work on the mass, converting kinetic energy to elastic potential energy. As kinetic energy \( KE=\frac{1}{2}mv^{2} \) decreases (because potential energy increases), velocity \( v \) decreases. So the correct description is that velocity decreases, acceleration increases, and net force increases.

Answer:

The velocity decreases,
the acceleration increases,
and the net force
increases.