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: Understand the forces opposing motion.
The block moves upward on the inclined plane. Therefore, two forces oppose its motion along the plane:
Gravitational component down the plane: \[ mg\sin\theta \] Frictional force down the plane: \[ \mu mg\cos\theta \] Hence, total retarding force is \[ F=mg\sin\theta+\mu mg\cos\theta \]

Step 2: Substitute the given values.
Given, \[ m=5\,\text{kg},\quad g=10\,\text{m s}^{-2},\quad \theta=45^\circ,\quad \mu=0.5 \] So, \[ F=5\times 10\sin45^\circ+0.5\times 5\times 10\cos45^\circ \] Since, \[ \sin45^\circ=\cos45^\circ=\frac{1}{\sqrt{2}} \] Therefore, \[ F=50\times \frac{1}{\sqrt{2}}+25\times \frac{1}{\sqrt{2}} \] \[ F=\frac{75}{\sqrt{2}}\,\text{N} \]

Step 3: Apply work-energy theorem.
The initial kinetic energy of the block is \[ K=100\,\text{J} \] When the block stops, its final kinetic energy becomes \[ 0 \] The work done against the retarding force is equal to the initial kinetic energy.
So, \[ Fs=100 \] \[ \frac{75}{\sqrt{2}}s=100 \]

Step 4: Calculate the distance.
Solving for \(s\), \[ s=\frac{100\sqrt{2}}{75} \] \[ s=\frac{4\sqrt{2}}{3}\,\text{m} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{4\sqrt{2}}{3}\,\text{m}} \]