Solution:
\[ K.E= \frac{1}{2}K\left( A^{2}-x^{2} \right)\]
\[ K.E= \frac{1}{2}\times 50\left( 10^{2}-5^{2} \right)/10^{4}= 0.19 J \]
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm–¹. The block is pulled to a distance of x = 10 cm from its equilibrium position at x = 0 cm on a frictionless surface from rest at t = 0. The kinetic energy of the block when it is 5 cm away from the mean position is
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