Question

A block of mass $10kg$ is attached to a relaxed string. Other end is attached to ceiling and block is released keeping the spring in vertical position. The spring is such that it has elongation of $10cm$ under a force of $100N$. the maximum velocity of block in subsequent motion is :-

• A. $1m/s^2$
• B. $1.5m/s^2$
• C. $\mathrm{2..0}m/s^2$
• D. $2.5m/s^2$

Answer 1

Answer: (A)

$1m/s^2$

First calculate spring constant
$P=kx$
$100=kx$
$100=k\times 0.1$
$k=1000N/m$
Now calculate elongation at equilibrium position
At equilibrium, net force $F=0$
$kx=mg$
$1000x=10\times 10$
$x=10cm=0.1m$
At $x=0.1m$, velocity will be maximum.
Using energy conservation
$mg\times 0.1=\frac{1}{2}\times 100\times (0.1{)}^2+\frac{1}{2}m{V}_{max}^2$
$10=5+\frac{1}{2}\times 10V_{max}^2$
$5=5V_{max}^2$
$V=1m/s^2$