A spring is hung vertically from a hook. A 500g mass is hung from the spring and the spring stretches 0.01m from equilibrium. If the mass is then pulled 0.02m MORE and released what will be the acceleration of the mass just as it is released? (show work)

See the explanation below.

Explanation:

By means of Hooke's second law, we can calculate the spring constant with the initial conditions of the problem.

where:

k = spring constant [N/m]

x = distance = 0.01 [m]

F = force or weight [N]

Weight can be calculated by means of the product of mass by gravitational acceleration.

m = mass = 500[g] = 0.5[kg]

Now the spring constant can be calculated:

We must now use the same Hooke's law to determine the new spring force when stretching it 0.03 [m]

Now we have to use Newton's second law to calculate acceleration. We must remember that Newton's Second Law tells us that the sum of forces is equal to the product of mass by acceleration.

where:

F = Fnew = 14.715[N]

m = mass = 500 [g] = 0.5 [kg]

a = acceleration [m/s²]

Now replacing:

answer; /// the above statement is (true);

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.