The drawing shows three situations in which a block is attached to a spring. The position labeled "0 m" represents the unstrained position of the spring. The block is moved from an initial position x0 to a final position xf, the magnitude of the displacement being denoted by the symbol s. Suppose the spring has a spring constant of k = 48.2 N/m. Using the data provided in the drawing, determine the total work done by the restoring force of the spring for each situation. In the case of zero put your result as "+0".

Respuesta :

Answer:

The drawing is attached as image and answers along with explanation is provided below.

Step-by-step explanation:

We are given a mass spring damper system to compute the total work done by the restoring force of the spring for 3 different cases.

The work done is given by

[tex]W=\frac{1}{2}k (x^{2}_{0}-x^{2}_{f})[/tex]

Where [tex]k=48.2[/tex] [tex]N/m[/tex] is the spring constant and [tex]x_{0}[/tex]  is the initial position and [tex]x_{f}[/tex] is the final position.

Case 1:

[tex]x_{0}=1[/tex]

[tex]x_{f}=3[/tex]

[tex]W=\frac{1}{2}48.2 (1^{2}-3^{2})[/tex]

[tex]W=\frac{1}{2}48.2 (1-9)[/tex]

[tex]W=\frac{1}{2}48.2 (-8)[/tex]

[tex]W=48.2 (-4)[/tex]

[tex]W=-192.8[/tex] [tex]joules[/tex]

Case 2:

[tex]x_{0}=-3[/tex]

[tex]x_{f}=1[/tex]

[tex]W=\frac{1}{2}48.2 ((-3)^{2}-1^{2})[/tex]

[tex]W=\frac{1}{2}48.2 (9-1)[/tex]

[tex]W=\frac{1}{2}48.2 (8)[/tex]

[tex]W=48.2 (4)[/tex]

[tex]W=192.8[/tex] [tex]joules[/tex]

Case 3:

[tex]x_{0}=-3[/tex]

[tex]x_{f}=3[/tex]

[tex]W=\frac{1}{2}48.2 ((-3)^{2}-3^{2})[/tex]

[tex]W=\frac{1}{2}48.2 (9-9)[/tex]

[tex]W=\frac{1}{2}48.2 (0)[/tex]

[tex]W=0[/tex] [tex]joules[/tex]

makes sense because initial and final positions are equal in magnitude but are opposite in sign therefore, net work done is zero.

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