A block slides down a slope of angle `theta` with constant velocity. It is then projected up with a velocity of `10ms^(-1), g=10ms^(-2)` & `theta` = 30°. The maximum distance it can go up the plane before coming to stop is

Options

A. 10 m
B. 5 m
C. 4 mg
D. 15 m

Answer

To solve the problem step by step, we will analyze the motion of the block on the slope and use the principles of physics to find the maximum distance it can travel up the slope before coming to a stop.

### Step 1: Understand the Forces Acting on the Block
When the block slides down the slope with constant velocity, the net force acting on it is zero. This means that the force of gravity acting down the slope is balanced by the frictional force acting up the slope.

- The gravitational force component acting down the slope is given by:
\[ F_{\text{gravity}} = mg \sin \theta \]
- The frictional force acting up the slope is given by:
\[ F_{\text{friction}} = \mu mg \cos \theta \]
- Since the block is moving with constant velocity, we have:
\[ mg \sin \theta = \mu mg \cos \theta \]

### Step 2: Find the Coefficient of Friction
From the equation \( mg \sin \theta = \mu mg \cos \theta \), we can cancel \( mg \) (assuming \( m \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]
Thus, the coefficient of friction \( \mu \) can be expressed as:
\[ \mu = \tan \theta \]

### Step 3: Analyze the Block When Projected Upward
When the block is projected upward with an initial velocity \( v = 10 \, \text{m/s} \), it will experience a deceleration due to gravity and friction. The total acceleration \( a \) when moving up the slope can be expressed as:
\[ a = g \sin \theta + \mu g \cos \theta \]
Substituting \( \mu \):
\[ a = g \sin \theta + \tan \theta \cdot g \cos \theta \]
This simplifies to:
\[ a = g \sin \theta + g \sin \theta = 2g \sin \theta \]

### Step 4: Calculate the Maximum Distance Up the Slope
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity when the block comes to rest),
- \( u = 10 \, \text{m/s} \) (initial velocity),
- \( a = -2g \sin \theta \) (deceleration),
- \( s \) is the distance traveled up the slope.

Rearranging gives:
\[ 0 = (10)^2 + 2(-2g \sin \theta)s \]
\[ 100 = 4g \sin \theta \cdot s \]
Thus:
\[ s = \frac{100}{4g \sin \theta} \]

### Step 5: Substitute Values
Given \( g = 10 \, \text{m/s}^2 \) and \( \theta = 30^\circ \) (where \( \sin 30^\circ = \frac{1}{2} \)):
\[ s = \frac{100}{4 \cdot 10 \cdot \frac{1}{2}} = \frac{100}{20} = 5 \, \text{m} \]

### Conclusion
The maximum distance the block can travel up the slope before coming to a stop is:
\[ \boxed{5 \, \text{m}} \]