Question

Two springs of force constants 500 N/m
and 250 N/m are stretched by same
force. The ratio of their respective
potential energies is​

Answer 1

Answer:

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Answer 2

Answer:

Answer

KE=

2

1

∗k∗x

2

k

1

=3000N/m,k

2

=1500N/m

KE

1

:KE

2

=k

1

:k

2

=3000:1500=2:1 for same x