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. `12^(@)`
B. `18^(@)`
C. `15^(@)`
D. `30^(@)`

Answer

To find the angle of the ramp with the horizontal, we can use the principles of energy conservation and trigonometry. Here’s a step-by-step solution:

### Step 1: Identify the parameters
- Mass of the block (m) = 8 kg (not needed for angle calculation)
- Length of the ramp (L) = 1.5 m
- Final velocity (v) = 2.5 m/s
- Initial velocity (u) = 0 m/s (block is released from rest)

### Step 2: Use the principle of conservation of energy
The potential energy (PE) at the top of the ramp will convert into kinetic energy (KE) at the bottom of the ramp.

The potential energy at the top is given by:
\[ PE = mgh \]
where \( h \) is the height of the ramp.

The kinetic energy at the bottom is given by:
\[ KE = \frac{1}{2} mv^2 \]

Setting the potential energy equal to the kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \]

### Step 3: Cancel mass (m) from both sides
Since mass (m) appears on both sides, we can cancel it out:
\[ gh = \frac{1}{2} v^2 \]

### Step 4: Solve for height (h)
Rearranging the equation gives:
\[ h = \frac{v^2}{2g} \]

Substituting \( v = 2.5 \, \text{m/s} \) and \( g = 9.81 \, \text{m/s}^2 \):
\[ h = \frac{(2.5)^2}{2 \times 9.81} \]
\[ h = \frac{6.25}{19.62} \]
\[ h \approx 0.318 \, \text{m} \]

### Step 5: Use trigonometry to find the angle (θ)
In a right triangle formed by the ramp:
- The length of the ramp (hypotenuse) = 1.5 m
- The height (opposite side) = 0.318 m

Using the sine function:
\[ \sin(\theta) = \frac{h}{L} \]
\[ \sin(\theta) = \frac{0.318}{1.5} \]
\[ \sin(\theta) \approx 0.212 \]

### Step 6: Calculate the angle (θ)
Now, take the inverse sine to find the angle:
\[ \theta = \sin^{-1}(0.212) \]
Calculating this gives:
\[ \theta \approx 12.2^\circ \]

### Final Answer
The angle of the ramp with the horizontal is approximately \( 12^\circ \).

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