Question

45 POINTS


PHYSICS PROBLEM INCLUDE WORK


(I know the answer is B 1 joule, I just need work and explanation)




An object with massm m is suspended at rest from a spring with a spring constant of 200 N/m. The length of the spring is 5.0 cm longer than its unstretched length L, as shown above. A person then exerts a force
on the object and stretches the spring an additional 5.0 cm. What is the total energy stored in the spring
at the new stretched length?

Answer 1

The elastic potential energy in the spring is 1 J

Explanation:

The elastic potential energy stored in a spring is given by:

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching/compression of the spring with respect to the unstretched length

For the spring in this problem, we have:

k = 200 N/m is its spring constant

The spring is stretched by 5.0 cm first and then by an additional 5.0 cm, so the total stretching is

$x=5.0 + 5.0 = 10.0 cm = 0.10 m$

Therefore, the elastic potential energy in the spring is:

$U=\frac{1}{2}(200)(0.10)^2=1 J$

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Answer 2

The Total  Energy stored  at the new stretched length in the  spring mass system  = 1  Joule

The Elastic Potential Energy Stored  in the  spring mass system  is given by equation (1)

E = (1/2) ........(1)

Where K is the Force Constant =  200 N/m

 The total extension of the spring = $x$

Total Extension  =   Initial Extension   + Final  Extension (cm)  

=  5 + 5 = 10 cm= 10/100 m = 0.1 m

E    = (1/2 )$\times$ 200

For more information please refer to the  link below  

https://brainly.com/question/12807194