A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is :

Solution

Physics sir · 4 steps

Follow the energy: gravity gives it, the rough 2 m patch takes some, and the spring stores what is left.

1. Energy from the incline. The block slides 10 m along a smooth 30° incline, so it drops h = 10 sin 30° = 5 m. With g = 10 m s⁻², it gains mgh = 5 × 10 × 5 = 250 J of kinetic energy.
2. Loss on rough patch. On the 2 m stretch with μ = 0.5, friction is μ mg = 0.5 × 5 × 10 = 25 N. It removes 25 × 2 = 50 J, leaving 200 J when the block reaches the spring.

   mgh - μ mg d = 1/2kx² ⇒ 250 - 50 = 1/2(100)x²

3. Solve for x. 200 = 50x², so x² = 4 and x = 2 m. At maximum compression the block is momentarily at rest, so all of the remaining energy is stored in the spring.
4. So. The maximum compression of the spring is 2 m, option (d).

The idea to keep. Write a single energy balance from start to the point of rest: energy gained from gravity minus work lost to friction equals energy stored in the spring. Friction acts only over the rough length.

Answer from the NTA final answer key. Solution by PrepWiser.