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The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the distance x the spring is stretched.
Figure1 of 1
A graph of force F as a function of distance x. The force is measured from 0 to 350 newtons on the vertical axis. The distance is measured from 0 to 8 centimeters on the horizontal axis. A graph curve starts from the origin and rises linearly to approximately 7 centimeters and 350 newtons.
Part A
How much work is needed to stretch this spring a distance of 2.0 cm, starting with it unstretched.
Express your answer in joules to two significant figures.


W =
nothing
J
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Part B
How much work is needed to stretch this spring from 5.0 cm to 7.0 cm?
Express your answer in joules to two significant figures.


W =
nothing
J
Request Answer
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Respuesta :

pstnonsonjoku

The elastic potential energy stored in the stretched spring is 1 J.

What is Hooke's law?

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

= 0.5 × 5000 × (2 × 10^-2)^2 =1 J

Learn more about elastic potential energy: https://brainly.com/question/156316

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