Question

A massless spring with a force constant $K=40N/m$ hangs vertically from the ceiling. A $0.2kg$ block is attached to the free end of the spring and held in such a position that the spring has its natural length and suddenly released. The maximum elastic strain energy stored in the spring is (take $g=10m/s^2$):

• A. $0.1J$
• B. $0.2J$
• C. $0.05J$
• D. $0.4J$

Answer 1

Answer: (B)

$0.2J$

The decrease in potential energy of $0.2kg$ mass will be gain in elastic energy of spring

$\Rightarrow (P.E)spring=\frac{1}{2}k{x}^2$...(i)

Decrease in potential energy of $0.2kg$ mass $=0.2gh$...(ii)

From equation (i) and (ii):

$\frac{1}{2}k{x}^2=.2gh$

$x=\frac{2mg}{k}$

Decrease in potential energy, $mgx=mg\times \left(\frac{2mg}{k}\right)$

$=2\frac{(mg{)}^2}{k}$

$=2\times \frac{(0.2{)}^2\times (10{)}^2}{40}$

$=\frac{4}{20}=0.2J$