A block is pushed up a rough inclined plane of 45°. If the time of descent is twice the time of ascent, the coefficient of friction is

Options

A. 0.6
B. 0.4
C. 0.5
D. 0.25

Answer

To solve the problem of a block being pushed up a rough inclined plane at an angle of 45°, where the time of descent is twice the time of ascent, we need to find the coefficient of friction. Let's break down the solution step by step.

### Step 1: Understanding the Forces Acting on the Block
When the block is on the inclined plane, the forces acting on it include:
- Gravitational force (mg) acting downwards.
- Normal force (N) acting perpendicular to the surface of the incline.
- Frictional force (f) acting opposite to the direction of motion.

### Step 2: Analyzing the Forces
For an inclined plane at an angle θ (which is 45° in this case):
- The component of gravitational force parallel to the incline: \( F_{\parallel} = mg \sin(θ) \)
- The component of gravitational force perpendicular to the incline: \( F_{\perpendicular} = mg \cos(θ) \)

Since θ = 45°, we have:
- \( F_{\parallel} = mg \sin(45°) = mg \frac{\sqrt{2}}{2} \)
- \( F_{\perpendicular} = mg \cos(45°) = mg \frac{\sqrt{2}}{2} \)

### Step 3: Normal Force and Friction
The normal force (N) is equal to the perpendicular component of the weight:
\[ N = mg \cos(45°) = mg \frac{\sqrt{2}}{2} \]

The frictional force (f) can be expressed as:
\[ f = \mu N = \mu mg \frac{\sqrt{2}}{2} \]
where \( \mu \) is the coefficient of friction.

### Step 4: Equations of Motion
Let \( t_a \) be the time of ascent and \( t_d \) be the time of descent. According to the problem, \( t_d = 2t_a \).

Using the equations of motion, we can express the acceleration during ascent and descent:
- For ascent (against gravity and friction):
\[ a_a = g \sin(45°) + \mu g \cos(45°) = g \frac{\sqrt{2}}{2} + \mu g \frac{\sqrt{2}}{2} \]
- For descent (with gravity and friction):
\[ a_d = g \sin(45°) - \mu g \cos(45°) = g \frac{\sqrt{2}}{2} - \mu g \frac{\sqrt{2}}{2} \]

### Step 5: Relating Time and Acceleration
Using the relationship between time, distance, and acceleration:
- For ascent:
\[ d = \frac{1}{2} a_a t_a^2 \]
- For descent:
\[ d = \frac{1}{2} a_d t_d^2 = \frac{1}{2} a_d (2t_a)^2 = 2 a_d t_a^2 \]

Setting the distances equal gives:
\[ \frac{1}{2} a_a t_a^2 = 2 a_d t_a^2 \]

### Step 6: Simplifying the Equation
Dividing both sides by \( t_a^2 \) and simplifying:
\[ \frac{1}{2} a_a = 2 a_d \]
\[ a_a = 4 a_d \]

### Step 7: Substituting for Accelerations
Substituting the expressions for \( a_a \) and \( a_d \):
\[ g \frac{\sqrt{2}}{2} + \mu g \frac{\sqrt{2}}{2} = 4 \left( g \frac{\sqrt{2}}{2} - \mu g \frac{\sqrt{2}}{2} \right) \]

### Step 8: Solving for the Coefficient of Friction
Now, simplifying the equation:
\[ g \frac{\sqrt{2}}{2} + \mu g \frac{\sqrt{2}}{2} = 4g \frac{\sqrt{2}}{2} - 4\mu g \frac{\sqrt{2}}{2} \]
\[ \mu g \frac{\sqrt{2}}{2} + 4\mu g \frac{\sqrt{2}}{2} = 4g \frac{\sqrt{2}}{2} - g \frac{\sqrt{2}}{2} \]
\[ 5\mu g \frac{\sqrt{2}}{2} = 3g \frac{\sqrt{2}}{2} \]
Dividing both sides by \( g \frac{\sqrt{2}}{2} \):
\[ 5\mu = 3 \]
\[ \mu = \frac{3}{5} \]

### Final Answer
The coefficient of friction \( \mu \) is \( \frac{3}{5} \).

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