Two springs of spring constant 1500 N-m and 3000 N-m respectively are stretched with the same force- They will have the potential energies in the ratio of1 - 21 - 42 - 14 - 1

Question

Toppr Admin · Accepted Answer

KE-12-x2217-k-x2217-x2k1-3000N-m-k2-1500N-mKE1-KE2-k1-k2-3000-1500-2-1 for same x

Two springs of spring constant $1500 N / m$ and $3000 N / m$ respectively are stretched with the same force. They will have the potential energies in the ratio of
$1 : 2$
$1 : 4$
$2 : 1$
$4 : 1$

$K E = \frac{1}{2} * k * x^{2}$
$k_{1} = 3000 N / m, k_{2} = 1500 N / m$
$K E_{1} : K E_{2} = k_{1} : k_{2} = 3000 : 1500 = 2 : 1$ for same $x$

Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have the potential energies in the ratio of1:21:42:14:1

KE=12∗k∗x2k1=3000N/m,k2=1500N/mKE1:KE2=k1:k2=3000:1500=2:1 for same x

Two springs of spring constant $1500 N / m$ and $3000 N / m$ respectively are stretched with the same force. They will have the potential energies in the ratio of
$1 : 2$
$1 : 4$
$2 : 1$
$4 : 1$

$K E = \frac{1}{2} * k * x^{2}$
$k_{1} = 3000 N / m, k_{2} = 1500 N / m$
$K E_{1} : K E_{2} = k_{1} : k_{2} = 3000 : 1500 = 2 : 1$ for same $x$