Question

A block of mass 5 kg is released as shown in the figure. Surface CD is rough with = 0.5, rest of all the surfaces are smooth. Find the maximum compression in the spring (initially spring is in its natural length.)

• A. 1.5 m
• B. 2.0 m
• C. 3.5 m
• D. 2.5 m

Answer 1

The Correct Option is B

To find the maximum compression in the spring, we need to analyze the motion of the block as it moves from the inclined plane to the spring. Let's go through the steps:

1. Calculate the potential energy at height:

   When the block is at the top of the incline at point A, it has potential energy due to its height above the ground. The height (h) can be calculated using the length of the incline (10 m) and the angle (60°).

   h = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m}

   The potential energy (PE) is given by:

   \text{PE} = mgh = 5 \times 9.8 \times 5\sqrt{3} \, \text{J}

2. Frictional force on rough surface:

   The rough surface CD causes energy loss due to friction. Its force of friction (f) is given by:

   f = \mu mg = 0.5 \times 5 \times 9.8 = 24.5 \, \text{N}

   The work done by friction (W) over distance CD is:

   W = f \times d = 24.5 \times x

   This is the energy lost.

3. Conservation of Energy:

   The initial potential energy is converted to spring potential energy and work done against friction:

   mgh = \frac{1}{2}kx^2 + f \times d

   Substituting values:

   245\sqrt{3} = \frac{1}{2} \times 100 \times x^2 + 24.5 \times x

   Simplifying:

   245 \times 1.732 = 50x^2 + 24.5x

   424.74 = 50x^2 + 24.5x

4. Solve the Quadratic Equation:

   50x^2 + 24.5x - 424.74 = 0

   Using the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, solve for x:

   a = 50, \, b = 24.5, \, c = -424.74

   Calculate b^2 - 4ac:

   24.5^2 + 4 \times 50 \times 424.74 = 600.25 + 84948 = 85548.25

   x = \frac{-24.5 \pm \sqrt{85548.25}}{100}

   The positive value of x is the maximum compression:

   x \approx 2.0 \, \text{m}

5. Conclusion:

   The maximum compression in the spring is 2.0 m.