The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8cm the potential energy stored in it is :
A) 4U
B) 8U
C) 16U
4) U/4
Rushikesh23:
Answer is 16U
Yess
Explanation
??
Since 1\2kx^2=PE of spring by given condition 1/2k(2)^2=U. Hence by solving it we get k=U/2 then by second condition PE=1/2(U/2)64 then we get 16 U
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Answers
Answered by
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PE=1/2kx^2
By first condition
U=1/2k(2)^2
Then we get
K=U/2
Then by second condition
PE=1/2kx^2
Substituting value of K & x=8
PE=1/2 (U/2) (8)^2
We get 16U
By first condition
U=1/2k(2)^2
Then we get
K=U/2
Then by second condition
PE=1/2kx^2
Substituting value of K & x=8
PE=1/2 (U/2) (8)^2
We get 16U
hi
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