Question

An ideal spring is hung vertically from the ceiling. When a 2kg mass hangs at rest the spring is elongated by 6cm from its relaxed length. A downward force is now applied to the mass to elongate the spring further by 10cm. When the spring is elongated by force, the work done by the spring is

• A. 36J
• B. -3.6J
• C. 4.26J
• D. -4.26J

Answer 1

The correct option is B

-3.6J

Step 1: Given data:
Mass of the spring m=2kg
Elongated length x₁ = 6 cm = 0.06m
Length of the spring after adding the mass x₂ =10 cm = 0.1m

Step 2: Formula

The force acting on the spring, after it elongated the force of gravity and Hooke’s law.
F= mg=k x ₁
Where m is the mass,
g is the acceleration due to gravity
k is the spring constant
Therefore
$k=\frac{mg}{x}=\frac{2\times 10}{0.6m}$
k=333.33N/m

Step 3: Work done by the spring when its extended by 6 cm is,
$W_1=-\frac{1}{2}K{x_1}^2$
$W_1=-\frac{1}{2}\times 333.33N\times {(6\times {10}^{-2})}^2$
$W_1=-0.599J$

Step 3: Work done by the spring when its extended by 10 cm is,
$x_2=10\times {10}^{-2}m+6\times {10}^{-2}m=16\times {10}^{-2}m$
$W_2=-\frac{1}{2}K{x^2}_2$
$W_2=-\frac{1}{2}\times 333.33N{(16\times {10}^{-2})}^2$
$W_2=-4.26J$

1. Work done W=W₂ - W₁
   W=-4.26J-(-0.599J)
   W=-3.6J

Hence, the correct option is B.