Question

(For a horizontal spring) determine the acceleration of a 0.250kg mass on the end of a 54.7N/m spring if it has been stretched 12cm from its equilibrium and released. ?

Answer 1

The acceleration of the mass will initially be `26.26 m/s^2.`

Explanation:

The typed and written versions of the question seem to disagree on the value of the spring constant. I will use 54.7 N/m.

Hooke's Law says that the force the spring exerts at its 0.12 m displacement will be `54.7 N/m * 0.12 m = 6.56 N`

Now let's take that force and use it in Newton's 2nd Law.
`F = m*a`
`6.56 N = 0.250 kg*a`

Solving for a
`a = (6.56 N)/(0.250 kg) = 26.26 m/s^2`

Note that the acceleration will immediately start decreasing because the force will decrease as the spring approaches its equilibrium position.

I hope this helps,
Steve