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  • A spring with spring constant  when stretched through 1cm, the potential energy is

A spring with spring constant k when stretched through 1cm, the potential energy is U. If it is stretched by 4cm. The potential energy will be 

Option: 1

4U


Option: 2

8U


Option: 3

16U


Option: 4

2U

 


Answers (1)

best_answer

The potential energy stored in the spring, i.e. stretched or compressed a certain distance x from its equilibrium position is given by the formula:

PE=\frac{1}{2} k x^2

where k is the spring constant.

Given that the potential energy of the spring is U when it is stretched by 1 cm (0.01 m), we can use this information to find the spring constant:

U=\frac{1}{2} \text k(0.01 \text m)^2

k=20000 \: \text U / \text m

Now we can use the spring constant to find the potential energy of the spring when it is stretched further by 4 cm (0.04 m):

PE=\frac{1}{2}k x^2=\frac{1}{2}(20000\: \text U / \text m) ( 0.04 \text m)^2 = 16 \text U

Therefore, the potential energy of the spring, when it is stretched by 4 cm, is 16 U.

 

Posted by

jitender.kumar

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