A block of mass 2.0 kg is pushed down an inclined plane of inclination `37^0` with a force of 20N cting parallelto the incline. It is found tht the block moves on the incline with an acceleration of `10 m/s^2`. If the blcok started from rest, find the work done a. by the applied force in the first second b. by the weight of the block in the first second and c. by the frictional force acting on the block in the first second. Take `g=10 m/s^2`.

Explanation

To solve the problem step by step, we will calculate the work done by the applied force, the weight of the block, and the frictional force during the first second of motion.

### Given Data:
- Mass of the block, \( m = 2.0 \, \text{kg} \)
- Inclination angle, \( \theta = 37^\circ \)
- Applied force, \( F = 20 \, \text{N} \)
- Acceleration, \( a = 10 \, \text{m/s}^2 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)
- Time, \( t = 1 \, \text{s} \)

### Step 1: Calculate the displacement in the first second
Using the formula for displacement under constant acceleration:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the block starts from rest, \( u = 0 \):
\[ s = 0 + \frac{1}{2} \times 10 \, \text{m/s}^2 \times (1 \, \text{s})^2 = \frac{1}{2} \times 10 \times 1 = 5 \, \text{m} \]

### Step 2: Calculate the work done by the applied force
The work done \( W \) by a force is given by:
\[ W = F \cdot s \cdot \cos(\theta) \]
Since the applied force is parallel to the incline, \( \theta = 0^\circ \) and \( \cos(0) = 1 \):
\[ W_{\text{applied}} = 20 \, \text{N} \times 5 \, \text{m} \times 1 = 100 \, \text{J} \]

### Step 3: Calculate the work done by the weight of the block
The weight of the block \( W = mg \):
\[ W = 2.0 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \]
The component of the weight acting down the incline is:
\[ W_{\text{down}} = mg \sin(\theta) = 20 \, \text{N} \cdot \sin(37^\circ) \]
Using \( \sin(37^\circ) \approx \frac{3}{5} \):
\[ W_{\text{down}} = 20 \, \text{N} \cdot \frac{3}{5} = 12 \, \text{N} \]
The work done by the weight in the first second is:
\[ W_{\text{weight}} = W_{\text{down}} \cdot s = 12 \, \text{N} \cdot 5 \, \text{m} = 60 \, \text{J} \]

### Step 4: Calculate the work done by the frictional force
Using the work-energy theorem:
\[ W_{\text{friction}} + W_{\text{applied}} + W_{\text{weight}} = \Delta KE \]
The change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 2.0 \, \text{kg} \times (10 \, \text{m/s})^2 = 100 \, \text{J} \]
Substituting the values:
\[ W_{\text{friction}} + 100 \, \text{J} + 60 \, \text{J} = 100 \, \text{J} \]
Thus,
\[ W_{\text{friction}} = 100 \, \text{J} - 100 \, \text{J} - 60 \, \text{J} = -60 \, \text{J} \]

### Summary of Results:
a. Work done by the applied force in the first second: **100 J**
b. Work done by the weight of the block in the first second: **60 J**
c. Work done by the frictional force acting on the block in the first second: **-60 J**