A 60 kg weight is dragged on a horizontal surface by a rope upto 2 metres. If coefficient of friction is `mu=0.5` the angle of rope with the surface is `60^(@)`and `g=9.8m//sec^(2),` then work done is

Options

A. 315.1 J
B. 15 J
C. 588 J
D. 197 J

Explanation

To solve the problem step by step, we will follow the physics principles involved in the scenario described.

### Step 1: Identify the Forces Acting on the Weight
The forces acting on the weight (mass \( m = 60 \, \text{kg} \)) are:
- The weight of the object, \( W = mg = 60 \times 9.8 = 588 \, \text{N} \) (acting downwards).
- The normal force \( R \) (acting upwards).
- The tension in the rope \( F \) (at an angle of \( 60^\circ \) to the horizontal).
- The frictional force \( F_k \) (acting opposite to the direction of motion).

### Step 2: Resolve the Tension into Components
The tension \( F \) can be resolved into horizontal and vertical components:
- Horizontal component: \( F \cos(60^\circ) = F \cdot \frac{1}{2} \)
- Vertical component: \( F \sin(60^\circ) = F \cdot \frac{\sqrt{3}}{2} \)

### Step 3: Apply Newton's Second Law in the Vertical Direction
Since the system is in equilibrium in the vertical direction:
\[ R + F \sin(60^\circ) = mg \]
Substituting the values:
\[ R + F \cdot \frac{\sqrt{3}}{2} = 588 \]
Thus, we can express \( R \) as:
\[ R = 588 - F \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Calculate the Frictional Force
The frictional force \( F_k \) is given by:
\[ F_k = \mu R \]
Substituting for \( R \):
\[ F_k = \mu \left(588 - F \cdot \frac{\sqrt{3}}{2}\right) \]
Given \( \mu = 0.5 \):
\[ F_k = 0.5 \left(588 - F \cdot \frac{\sqrt{3}}{2}\right) \]

### Step 5: Apply Newton's Second Law in the Horizontal Direction
In the horizontal direction, the frictional force equals the horizontal component of the tension:
\[ F_k = F \cos(60^\circ) \]
Substituting for \( \cos(60^\circ) \):
\[ F_k = F \cdot \frac{1}{2} \]

### Step 6: Set Up the Equation
Equating the two expressions for \( F_k \):
\[ 0.5 \left(588 - F \cdot \frac{\sqrt{3}}{2}\right) = F \cdot \frac{1}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 588 - F \cdot \frac{\sqrt{3}}{2} = F \]
Rearranging gives:
\[ 588 = F + F \cdot \frac{\sqrt{3}}{2} \]
Factoring out \( F \):
\[ 588 = F \left(1 + \frac{\sqrt{3}}{2}\right) \]
Thus:
\[ F = \frac{588}{1 + \frac{\sqrt{3}}{2}} \]

### Step 7: Calculate the Value of \( F \)
Calculating the denominator:
\[ 1 + \frac{\sqrt{3}}{2} \approx 1 + 0.866 \approx 1.866 \]
So:
\[ F \approx \frac{588}{1.866} \approx 315.8 \, \text{N} \]

### Step 8: Calculate the Frictional Force
Now substituting \( F \) back to find \( F_k \):
\[ F_k = F \cdot \frac{1}{2} = 315.8 \cdot \frac{1}{2} \approx 157.9 \, \text{N} \]

### Step 9: Calculate the Work Done
The work done against friction over a distance \( d = 2 \, \text{m} \):
\[ \text{Work} = -F_k \cdot d = -157.9 \cdot 2 = -315.8 \, \text{J} \]
The magnitude of work done is:
\[ \text{Work} = 315.8 \, \text{J} \]

### Final Answer
The work done is approximately \( 315.8 \, \text{J} \).

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