Question

A block of mass \(5\,\text{kg}\) starts up a \(45^\circ\) incline with an initial kinetic energy of \(100\,\text{J}\). The coefficient of friction is \(0.5\). The distance it travels before coming to rest is:

• A. \( \dfrac{3}{\sqrt{2}}\,\text{m} \)
• B. \( 2\sqrt{2}\,\text{m} \)
• C. \( \dfrac{6\sqrt{2}}{2}\,\text{m} \)
• D. \( \dfrac{4\sqrt{2}}{3}\,\text{m} \)

Hint:

On an incline, always resolve gravity and friction components.
Use work-energy theorem for deceleration problems: \( \text{K.E. lost} = \text{work done against forces} \)
Trigonometric values for \(45^\circ\) often simplify expressions with square roots.

Answer 1

The Correct Option is D

Step 1: Use work-energy theorem. All kinetic energy is used against friction and gravity: \[ \text{Work done} = \text{Loss of kinetic energy} = 100\,\text{J} \] Step 2: Total resisting force = \( mg\sin\theta + \mu mg\cos\theta \) Given: - \( m = 5\,\text{kg},\ \mu = 0.5,\ \theta = 45^\circ \) - \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \) So: \[ F = 5 \cdot 10 \left( \frac{1}{\sqrt{2}} + 0.5 \cdot \frac{1}{\sqrt{2}} \right) = 50 \cdot \frac{3}{2\sqrt{2}} = \frac{75}{\sqrt{2}} \] Step 3: Let stopping distance be \(s\): \[ \frac{75}{\sqrt{2}} \cdot s = 100 \Rightarrow s = \frac{100 \sqrt{2}}{75} = \frac{4\sqrt{2}}{3} \]