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. 
 

Answer 2

To solve this problem, let's first analyze the situation using the principles of energy conservation. The energy transformations involved are gravitational potential energy, kinetic energy, and elastic potential energy of the spring.

Given:

• Force required to compress spring \( 1 \) m: \( 100 \) N
• Angle of incline: \( 30^\circ \)
• Mass of block \( M \): \( 10 \) kg
• Compression of spring: \( 2 \) m
• Gravitational acceleration \( g \): \( 10 \) m/s\(^2\)

Step 1: Determine the spring constant \( k \).

From Hooke's Law, \( F = kx \), for \( x = 1 \) m and \( F = 100 \) N:

\( k = \frac{F}{x} = \frac{100}{1} = 100 \) N/m

Step 2: Calculate energy at different stages.

When the block is at rest at the top, it only has potential energy.

Potential Energy (PE_i) = \( mgh \)

The vertical height \( h \) from the top of the incline can be found using trigonometry:

\( h = d \sin(30^\circ) \) where \( d \) is the distance traveled along the incline.

The spring is compressed 2 m, so the block travels 2 m along the incline (since the spring is at the bottom of the incline).

The height becomes: \( h = 2 \sin(30^\circ) = 1 \) m

Initial potential energy:

\( PE_i = mgh = 10 \times 10 \times 1 = 100 \) J

At the maximum compression, all of the gravitational potential energy has transformed into spring potential energy.

Spring Potential Energy (SPE) = \(\frac{1}{2}kx^2\)

For \( x = 2 \):

\( SPE = \frac{1}{2} \times 100 \times (2)^2 = 200 \) J

Step 3: Apply the Conservation of Energy principle.

The initial gravitational potential energy is transformed into kinetic energy and spring potential energy:

\( PE_i = KE + SPE \)

Solving for kinetic energy (KE):

\( 100 = KE + 200 \)

\( KE = 100 - 200 = 0 \)

Energy is conserved along the path since the incline is frictionless, hence we re-evaluate kinetic energy at the point just before touching the spring:

Re-interpret: As the block descends, it gains speed, converting potential to kinetic energy. When not yet touching the spring:

\( PE_{\text{lost}} = KE_{\text{gained}}\) Thus, \( 100 - 0 = KE + 0 \) confirms error; the descent naturally builds :

KE = \( 100 \).

\(\frac{1}{2}mv^2 = 100\)

Solving for \( v \):

\( \frac{1}{2} \times 10 \times v^2 = 100 \)

\( 5v^2 = 100 \)

\( v^2 = 20 \)

\( v = \sqrt{20} \) m/s

Thus, the speed of the mass just before it touches the spring is \( \sqrt{20} \) m/s.