Question

An ideal spring is hung vertically from the ceiling. When a $2.0kg$ mass hangs at rest from the spring is extended $6.0cm$ from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional $10cm$. While the spring is being extended by the force, the work done by the spring is:

• A. $-3.3J$
  
• B. $3.3J$
  
• C. $-1.67J$
  
• D. $3.6J$
  

Answer 1

The correct option is C $-1.67J$

When the spring is extended by 6 cm,in this equilibrium position forces acting on it are force of gravity and Hooke's force of restoration .

$F=mg=kx$

$k=\frac{mg}{x}=\frac{2\times 10}{6\times {10}^{-2}}$

$=\frac{1}{3}\times {10}^3$

When the spring is established by another 10 cm,the restoring force does a work equal to the potential energy of the spring:

$w=-\frac{1}{2}\times k\times x^2$

$=-\frac{1}{2}\times \frac{1}{3}\times {10}^3\times (10\times {10}^{-2}{)}^2$

$=-\frac{1}{6}\times {10}^3\times {10}^{-2}$

$-\frac{1}{6}\times 10=-1.67J$