- #1
mbrmbrg
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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 [tex]W=-(k\frac{x_f^2}{2}-k\frac{x_i^2}{2})[/tex], 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?
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 [tex]W=-(k\frac{x_f^2}{2}-k\frac{x_i^2}{2})[/tex], 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?