Question

A 500 gram mass is attached to horizontal spring, the spring is initially stretched by 5 cm & mass is released from there. If after $1/2$second mass crosses mean position, then find maximum kinetic energy of block.

• A. 3.15 mJ
• B. 6.25 mJ
• C. 4.25 mJ
• D. 7.45 mJ

Answer 1

The correct option is B 6.25 mJ

Here, particle takes $1/2$ second from extreme to mean position, hence
$1/2=T/4⟹T=2$ seconds

$V_{max}=A.w=5\times {10}^{-2}\times \frac{2\pi }{T}$
$=5\times {10}^{-2}\times \pi =\frac{5\pi }{100}$

$K{E}_{max}=\frac{1}{2}m{V}_{max}^2=\frac{1}{2}.\frac{1}{2}\times (\frac{5\pi }{100}{)}^2$
$=\frac{25{\pi }^2}{4\times 100\times 100}J$

$K{E}_{max}\approx 6.25mJ$