Solving for Block Position After 3.3J Work on Spring-Block System

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To determine the block's position after applying 3.3 J of work on the spring-block system, the spring constant was calculated to be -4450 N/m using the force of 89 N at x = -2.0 cm. The work-energy principle was applied, leading to the equation W = -(k(x_f^2/2) - k(x_i^2/2)), which yielded the positions of +/- 4.3 cm for the block. The work done by the person on the spring translates into stored elastic potential energy, while the net work is zero since there is no change in kinetic energy. This analysis confirms that the work done by the applied force equals the work done by the spring force in the opposite direction.
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In the figure below, we must apply a force of magnitude 89 N to hold the block stationary at x = -2.0 cm. From that position we then slowly move the block so that our force does +3.3 J of work on the spring-block system; the block is then again stationary. What is the block's position? (There are two answers.)

The figure is a block resting on a horizontal surface, attatched to a horizontal spring.

-our force=spring force=-kd
so -89J=-k(-0.02m)==>k=-4450N/m

I then said that W=-(k\frac{x_f^2}{2}-k\frac{x_i^2}{2}), subbed in W=3.3 J, k=-4450N/m, x_i=-0.02m, and got the right answer (which is +/- 4.3cm).
But why did that equation work? 3.3 J is our work, not the spring's?
 
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mbrmbrg said:
In the figure below, we must apply a force of magnitude 89 N to hold the block stationary at x = -2.0 cm. From that position we then slowly move the block so that our force does +3.3 J of work on the spring-block system; the block is then again stationary. What is the block's position? (There are two answers.)

The figure is a block resting on a horizontal surface, attatched to a horizontal spring.

-our force=spring force=-kd
so -89J=-k(-0.02m)==>k=-4450N/m

I then said that W=-(k\frac{x_f^2}{2}-k\frac{x_i^2}{2}), subbed in W=3.3 J, k=-4450N/m, x_i=-0.02m, and got the right answer (which is +/- 4.3cm).
But why did that equation work? 3.3 J is our work, not the spring's?
Yes, that is the work done by the person on the spring, The work results in stored elastic potential energy in the spring. If you look at the work energy theorem
W_{net} = \Delta K.E. and since there is no KE change,
W_{net} = 0. But the net work is the sum of the work by the conservative forces (spring force) and non conservative forces (the 'we' pushing force). They must therefore be equal and opposite. We do so many joules of work, and the spring force does the same in the other direction.
Alternatively, you can use the total work energy method
KE_i + PE_{gravity}_i +PE_{spring}_i = KE_f + PE_{gravity}_f +PE_{spring}_f + W_{n.c.}
and since there is no kinetic or gravitational potential energy change, you get the same result for the work done by "we".
 
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