Question

150 N shows an extension of 1.5 m. This spring is arranged as shown in figure. A block of mass 10 kg is released at rest on a frictionless incline. It slides down and compresses the spring by 3 m before coming to rest momentarily. Distance by which the block slides before hitting the spring is

• A. 4 m
• B. 6 m
• C. 9 m
• D. 12 m

Answer 1

Answer: (B)

6 m

Answer 2

The problem involves the application of Hooke's Law and the principle of conservation of mechanical energy.

1. Calculate the spring constant (k):

   The problem states that a force of 150 N causes an extension of 1.5 m in the spring. According to Hooke's Law, $F = kx$, where F is the force, k is the spring constant, and x is the extension/compression. Given $F = 150 \text{ N}$ and $x = 1.5 \text{ m}$.

   $k = \frac{F}{x} = \frac{150 \text{ N}}{1.5 \text{ m}} = 100 \text{ N/m}$

2. Apply the principle of conservation of mechanical energy:

   The block is released from rest on a frictionless incline. It slides down, first for a distance 'd' before hitting the spring, and then compresses the spring by 3 m before coming to rest momentarily. Since there is no friction, the total mechanical energy of the block-spring-Earth system is conserved.

   Let's consider the initial state when the block is released (at rest) and the final state when the spring is maximally compressed and the block is momentarily at rest.

   • Initial Energy: The block is at rest, so its initial kinetic energy is 0. The spring is not yet compressed, so its initial elastic potential energy is 0. We'll consider the initial gravitational potential energy relative to the final lowest point of the block.
   • Final Energy: The block is momentarily at rest, so its final kinetic energy is 0. The spring is compressed by $x_{spring} = 3 \text{ m}$, so its final elastic potential energy is $\frac{1}{2}kx_{spring}^2$. We'll set the final gravitational potential energy to 0.

   The total distance slid along the incline from the release point to the point of maximum compression is $D = d + x_{spring} = d + 3 \text{ m}$. The vertical height descended by the block, $\Delta h$, is related to the distance slid along the incline by $\Delta h = D \sin\theta$. Given $\theta = 30^\circ$, so $\sin\theta = \sin(30^\circ) = \frac{1}{2}$. Therefore, $\Delta h = (d + 3) \times \frac{1}{2}$.

   The gravitational potential energy lost by the block is $mg\Delta h$. The elastic potential energy gained by the spring is $\frac{1}{2}kx_{spring}^2$.

   By conservation of mechanical energy: Loss in gravitational potential energy = Gain in elastic potential energy $mg\Delta h = \frac{1}{2}kx_{spring}^2$ $mg(d + x_{spring})\sin\theta = \frac{1}{2}kx_{spring}^2$

   Substitute the known values: Mass of the block, $m = 10 \text{ kg}$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$ (standard assumption for JEE/NEET unless specified otherwise) Spring constant, $k = 100 \text{ N/m}$ Spring compression, $x_{spring} = 3 \text{ m}$ Angle of inclination, $\theta = 30^\circ$

   $10 \times 10 \times (d + 3) \times \sin(30^\circ) = \frac{1}{2} \times 100 \times (3)^2$ $100 \times (d + 3) \times \frac{1}{2} = 50 \times 9$ $50 \times (d + 3) = 450$ Divide both sides by 50: $d + 3 = \frac{450}{50}$ $d + 3 = 9$ $d = 9 - 3$ $d = 6 \text{ m}$

Thus, the distance by which the block slides before hitting the spring is 6 m.