An 8 kg block of ice, released from rest at the top of a 1.5 m long smooth ramp, slides down and falls with a velocity `2.5ms^(-1)`. Find angle of the ramp with horizontal.

Options

A. (a)`12^(@)`
B. (b)`18^(@)`
C. (c)`15^(@)`
D. (d)`30^(@)`

Answer

To solve the problem step by step, we will follow the principles of kinematics and dynamics.

### Step 1: Understand the Problem
We have an 8 kg block of ice that slides down a smooth ramp of length 1.5 m and reaches a velocity of 2.5 m/s at the bottom. We need to find the angle of the ramp with the horizontal.

### Step 2: Use the Kinematic Equation
We can use the kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and distance (s):
\[ v^2 = u^2 + 2as \]
Given:
- Initial velocity, \( u = 0 \) (released from rest)
- Final velocity, \( v = 2.5 \, \text{m/s} \)
- Distance, \( s = 1.5 \, \text{m} \)

Substituting the values into the equation:
\[ (2.5)^2 = 0 + 2a(1.5) \]
\[ 6.25 = 3a \]

### Step 3: Solve for Acceleration (a)
Now, we can solve for acceleration \( a \):
\[ a = \frac{6.25}{3} \]
\[ a \approx 2.0833 \, \text{m/s}^2 \]

### Step 4: Relate Acceleration to the Angle of the Ramp
On the ramp, the component of gravitational force causing the acceleration down the ramp is given by:
\[ ma = mg \sin \theta \]
Where:
- \( m \) is the mass of the block (which cancels out),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

From the equation:
\[ a = g \sin \theta \]
Substituting the values:
\[ 2.0833 = 9.8 \sin \theta \]

### Step 5: Solve for \( \sin \theta \)
Rearranging gives:
\[ \sin \theta = \frac{2.0833}{9.8} \]
\[ \sin \theta \approx 0.2124 \]

### Step 6: Find the Angle \( \theta \)
To find the angle \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}(0.2124) \]
Calculating this gives:
\[ \theta \approx 12.2^\circ \]

### Conclusion
Thus, the angle of the ramp with the horizontal is approximately \( 12^\circ \).