In the arrangement of the first figure, we gradually pull the block from x = 0 to x = +3.0 cm, where it is stationary. The second figuregives the work that our force does on the block. The scale of the figure's vertical axis is set by Ws = 1.0 J. We then pull the block out to x= +6.0 cm and release it from rest. How much work does the spring do on the block when the block moves from x;= +6.0 cm to (a) x =+4.0 cm, (b) x = -1.0 cm, and (c) x = -6.0 cm?x=0Ę = 000000000x positiveF negativeumurW (J)W₁01(a)(b)(c)2-Blockattachedto springرx (cm)x negativeF, positive3xXX

Name: In the arrangement of the first figure, we gradually pull the block f
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Description: In the arrangement of the first figure, we gradually pull the block from x = 0 to x = +3.0 cm, where it is stationary. The second figuregives the work that our force does on the block. The scale of the figure's vertical axis is set by Ws = 1.0 J. We

Answered: In the arrangement of the first figure,…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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In the arrangement of the first figure, we gradually pull the block from x = 0 to x = +3.0 cm, where it is stationary. The second figure
gives the work that our force does on the block. The scale of the figure's vertical axis is set by Ws = 1.0 J. We then pull the block out to x
= +6.0 cm and release it from rest. How much work does the spring do on the block when the block moves from x;= +6.0 cm to (a) x =
+4.0 cm, (b) x = -1.0 cm, and (c) x = -6.0 cm?
x=0
Ę = 0
00000000
x positive
F negative
umur
W (J)
W₁
0
1
(a)
(b)
(c)
2
-Block
attached
to spring
ر
x (cm)
x negative
F, positive
3
x
X
X

Expert Solution

Step 1

On the W axis, 10 small boxes has a total value of $W_{s} = 1.0 J V a l u e o f e a c h s m a l l b o x i n v e r t i c a l a x i s = \frac{1.0}{10} = 0.1 J T h e r e f o r e, t h e w o r k d o n e i n e x t e n d i n g t h e s p r i n g t o x = + 3 c m i s W = 0.9 J$

The work graph has vertical axis of energy and horizontal axis as expansion. So, the energy of the spring by stretching it up to x=+3.0 cm will be

$E_{x = 3} = 0.9 J \frac{1}{2} k x^{2} = 0.9 \frac{1}{2} k {(0.03)}^{2} = 0.9 k = \frac{0.9 \times 2}{{(0.03)}^{2}} k = 2000$

Here, k is the spring constant.

Work done by the spring W= Change in potential energy = $\frac{1}{2} k x_{i}^{2} - \frac{1}{2} k x^{2}$

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