Question

A block of mass \(5 \, \text{kg}\) starts up a \(45^\circ\) incline plane with initial kinetic energy of \(100 \, \text{J}\). If the coefficient of friction between block and plane is \(0.5\), then the distance covered by the block before it stops is \((g=10\,\text{m s}^{-2})\):

• A. \(\dfrac{4\sqrt{2}}{3}\,\text{m}\)
• B. \(\dfrac{3}{\sqrt{2}}\,\text{m}\)
• C. \(2\sqrt{2}\,\text{m}\)
• D. \(\dfrac{6\sqrt{2}}{5}\,\text{m}\)

Hint:

When a body moves upward on a rough inclined plane, both \(mg\sin\theta\) and friction \(\mu mg\cos\theta\) act down the plane and oppose the motion.

Answer 1

The Correct Option is A

Step 1: Calculate the net retarding force along the incline.
Both gravity component and kinetic friction oppose upward motion: \[ F_{ret} = mg(\sin45^\circ + \mu\cos45^\circ) = 5\times10\times\frac{1}{\sqrt{2}}(1+0.5) = \frac{75}{\sqrt{2}}\,\text{N} \]

Step 2: Use work-energy theorem.
Initial KE = 100 J, final KE = 0: \[ d = \frac{100}{F_{ret}} = \frac{100\sqrt{2}}{75} = \frac{4\sqrt{2}}{3}\,\text{m} \] \[ \boxed{\dfrac{4\sqrt{2}}{3}\,\text{m}} \]