Question

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 :

• A. \(\sqrt{6} \, \text{m}\)
• B. \(2 \, \text{m}\)
• C. \(1 \, \text{m}\)
• D. \(\sqrt{5} \, \text{m}\)

Answer 1

The Correct Option is B

To solve this problem, we need to calculate the maximum compression in the spring when the block hits it. We'll use energy conservation principles, factoring in gravitational potential energy, work done by friction, and spring potential energy.

1. Initially, the block is released from rest, meaning its initial kinetic energy is zero. It starts with gravitational potential energy at the incline's top, given by: \(U_g = mgh\), where \(m = 5 \, \text{kg}\), \(g = 10 \, \text{m/s}^2\) (assuming near the Earth's surface), and \(h = 10 \sin 30^\circ \, \text{m}\).
2. Calculate the height: \(h = 10 \times \frac{1}{2} = 5 \, \text{m}\)
3. Substitute into the potential energy formula: \(U_g = 5 \times 10 \times 5 = 250 \, \text{J}\)
4. The block slides down the incline and across a rough surface (friction coefficient \(\mu = 0.5\)) for \(2 \, \text{m}\). Calculate the work done by friction: \(W_f = \mu mg d = 0.5 \times 5 \times 10 \times 2 = 50 \, \text{J}\)
5. By energy conservation, the gravitational potential energy is converted into spring potential energy and work done against friction: \(U_g = \frac{1}{2} k x^2 + W_f\)
6. Substitute the known values: \(250 = \frac{1}{2} \times 100 \times x^2 + 50\)
7. Solve for \(x\):
   • First, simplify the equation: \(250 = 50 + 50 x^2\)
   • Re-arrange to solve for \(x^2\): \(200 = 50 x^2\)
   • Divide both sides by 50: \(x^2 = 4\)
   • Solve for \(x\): \(x = 2\)

Therefore, the maximum compression in the spring when the block hits the spring is \(2 \, \text{m}\).

Answer 2

Using the work-energy theorem, the total work done by all forces is equal to the change in kinetic energy (\(\Delta KE\)) of the system. Here:
\[w_g + w_{Fr} + w_s = \Delta KE.\]
Substituting the given values:
\[5 \times 10 \times 5 - 0.5 \times 5 \times 10 \times x - \frac{1}{2} Kx^2 = 0 - 0.\]
Simplifying:
\[250 - 25x - 50x^2 = 0.\]
Rewriting:
\[2x^2 + x - 10 = 0.\]
Solving this quadratic equation gives:
\[x = 2.\]
Thus, the maximum compression in the spring is \(x = 2\) m.