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Fundamentals Of Physics
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Short Answer

Figure 7-37 gives spring force Fx versus position x for the spring–block arrangement of Fig . 7-10 . The scale is set by Fx=160.0 N. We release the block at x=12 cm. How much work does the spring do on the block when the block moves from xi=+8.0 cm to (a) x= +5.0 cm, (b) x=-5.0 cm, (c) x=-8.0 cm, and (d) x=-10.0 cm ?

  1. The work done by the spring on the block when the block moves from xi=+8.0 cm to xf=5.0 cm is W=16 J
  2. The work done by the spring on the block when the block moves from xi=+8.0 cm to xf=-5.0 cm is W=16 J
  3. The work done by the spring on the block when the block moves from xi=+8.0 cm to xf=-8.0 cm is W=0 J
  4. The work done by the spring on the block when the block moves from xi=+8.0 cm to xf=-10.0 cm is W=-14 J
See the step by step solution

Step by Step Solution

Step 1: Understand the concept

We can use the equation of Hook’s law to calculate the spring constant from the graph, hence, we can calculate the work done in each case using the equation of work done by the spring.

Formulae:

F=-kxW=12k(xi2-xf2)

Given:

  1. The graph of Fx vs x
  2. The scale of graph is set at, Fs=160.0 N

Step 2: Calculate the value of spring constant

The graph of Fx vs x is given, so according to Hook’s law, the slope of this graph will give us the spring constant of the spring used.

From this graph, we get

slope=yx=80 N-1 cm=-80 Ncm=-8×103 N/m.

As per Hook’s law, we have

F=-kx

So,k=-Fx

Hence, the spring constant becomes,

k=8×103 N/m.

Step 3: a) Calculate the work does the spring do on the block when the block moves from xi=+8.0 cm to xf=5.0 cm

The equation for work done by the spring is,

W=12kxi2-xf2

For this part, xi=0.08 m and xf=0.05 m .

So,

W=12×8×103N.m0.080 m2-0.050 m2W=15.6 J16 J

Therefore, work done is, 16 J.

Step 4: b) Calculate the work does the spring do on the block when the block moves from xi=+8.0 cm to xf=-5.0 cm  

The equation for work done by spring is

W=12kxi2-xf2

For this part, xi=0.08 m and xf=-0.05 m. So,

W=12×8×103N.m0.080 m2--0.050 m2W=15.6 J16 J

Therefore, work done is, 16 J.

Step 5: c) Calculate the work does the spring do on the block when the block moves from xi=+8.0 cm to xf=-8.0 cm 

The equation for work done by spring is

W=12kxi2-xf2

For this part, xi=0.08 m and xf=-0.08 m. So,

W=12×8×103N.m0.080 m2--0.080 m2W=0 J

Therefore, work done is, 0 J.

Step 6: d) Calculate the work does the spring do on the block when the block moves from xi=+8.0 cm to xf=-10.0 cm 

We have, the equation for work done by spring as,

W=12kxi2-xf2

For this part, xi=0.08 m and xf=-0.10 m. So,

W=12×8×103N.m0.080 m2--0.10 m2W=-14.4 J-14 J

Therefore, work done is, -14 J.

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