Question

An ideal massless spring \( S \) can be compressed \( 1 \) m by a force of \( 100 \) N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined at \( 30^\circ \) to the horizontal. A \( 10 \) kg block \( M \) is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by \( 2 \) m. If \( g = 10 \) m/s\( ^2 \), what is the speed of the mass just before it touches the spring? 
 

• A. \( \sqrt{20} \) m/s
• B. \( \sqrt{30} \) m/s
• C. \( \sqrt{10} \) m/s
• D. \( \sqrt{40} \) m/s

Hint:

For spring problems, use conservation of energy: \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 + mgh \).

Answer 1

The Correct Option is A

Step 1: {Determine spring constant} 
Hooke’s Law: \[ F = kx \] \[ k = \frac{F}{x} = \frac{100}{1} = 100 { N/m} \] Step 2: {Use energy conservation} 
\[ \frac{1}{2} m v^2 + mgh = \frac{1}{2} kx_{\max}^2 \] \[ v = \sqrt{\frac{kx_{\max}^2}{m} - 2gh} \] Substituting values: \[ v = \sqrt{\frac{(100)(2)^2}{10} - (2)(10)(2/2)} \] \[ = \sqrt{20} { m/s} \] Thus, the correct answer is  \( \sqrt{20} \) m/s.