Question

A vertical spring of spring constant \(24Nm^{-1}\) is fixed on a table. A ball of mass \(0.5\,kg\) at a height \(2\,m\) above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance of \(50\,cm\). The net work done in the process is:

• A. \(6.5\,J\)
• B. \(10.5\,J\)
• C. \(12.5\,J\)
• D. \(9.5\,J\)

Hint:

For falling body on spring, include both free fall distance and compression distance while calculating work done by gravity.

Answer 1

The Correct Option is D

Step 1: Identify the total downward displacement.
The ball first falls \(2\,m\) before touching the spring, and then compresses the spring by \(50\,cm = 0.5\,m\).
\[ s = 2 + 0.5 = 2.5\,m \]

Step 2: Work done by gravity.
\[ W_g = mgs \]
Taking \(g = 10\,ms^{-2}\),
\[ W_g = 0.5 \times 10 \times 2.5 \]
\[ W_g = 12.5\,J \]

Step 3: Work done against spring force.
Energy stored in compressed spring is:
\[ W_s = \frac{1}{2}kx^2 \]
\[ W_s = \frac{1}{2} \times 24 \times (0.5)^2 \]

Step 4: Calculate spring work.
\[ W_s = 12 \times 0.25 \]
\[ W_s = 3\,J \]

Step 5: Net work done.
Net work done on the ball is work done by gravity minus work stored in spring.
\[ W_{\text{net}} = W_g - W_s \]
\[ W_{\text{net}} = 12.5 - 3 \]

Step 6: Final conclusion.
\[ W_{\text{net}} = 9.5\,J \]
\[ \boxed{9.5\,J} \] Hence, correct answer is option (D).