A body of mass 0.5 kg is taken up an inclined plane of length 10 m and height 5 m and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.1. The work done by the frictional force over the round trip is

Options

A. 5 J
B. `5sqrt3J`
C. `-5J`
D. `-5sqrt3J`

Explanation

To solve the problem step by step, we will calculate the work done by the frictional force over the round trip of the body on the inclined plane.

### Step 1: Identify the parameters
- Mass of the body, \( m = 0.5 \, \text{kg} \)
- Length of the inclined plane, \( L = 10 \, \text{m} \)
- Height of the inclined plane, \( h = 5 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.1 \)

### Step 2: Calculate the angle of inclination
Using the sine function:
\[ \sin \theta = \frac{h}{L} = \frac{5}{10} = 0.5 \]
Thus, \( \theta = 30^\circ \).

### Step 3: Calculate the normal force
The normal force \( N \) acting on the body when it is on the inclined plane can be calculated using:
\[ N = mg \cos \theta \]
Where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).
\[ N = 0.5 \times 10 \times \cos(30^\circ) = 0.5 \times 10 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3} \, \text{N} \]

### Step 4: Calculate the frictional force
The frictional force \( f \) can be calculated using:
\[ f = \mu N = 0.1 \times 2.5\sqrt{3} = 0.25\sqrt{3} \, \text{N} \]

### Step 5: Calculate the work done by friction for one trip
The work done by friction \( W \) when the body moves up the incline is given by:
\[ W = f \cdot d \cdot \cos(180^\circ) \]
Where \( d \) is the distance traveled (10 m) and \( \cos(180^\circ) = -1 \):
\[ W_{\text{up}} = 0.25\sqrt{3} \times 10 \times (-1) = -2.5\sqrt{3} \, \text{J} \]

### Step 6: Calculate the work done by friction for the downward trip
When the body slides down, the frictional force acts in the opposite direction:
\[ W_{\text{down}} = f \cdot d \cdot \cos(0^\circ) \]
Where \( \cos(0^\circ) = 1 \):
\[ W_{\text{down}} = 0.25\sqrt{3} \times 10 \times 1 = 2.5\sqrt{3} \, \text{J} \]

### Step 7: Calculate the total work done by friction over the round trip
The total work done by friction over the round trip is:
\[ W_{\text{total}} = W_{\text{up}} + W_{\text{down}} = -2.5\sqrt{3} + 2.5\sqrt{3} = 0 \, \text{J} \]

However, since the question asks for the work done by friction over the entire round trip, we need to consider the absolute value of the work done during both trips:
\[ W_{\text{friction}} = 2 \times |W_{\text{up}}| = 2 \times 2.5\sqrt{3} = 5\sqrt{3} \, \text{J} \]
But since the work done by friction is negative (indicating energy loss), we write:
\[ W_{\text{friction}} = -5\sqrt{3} \, \text{J} \]

### Final Answer
The work done by the frictional force over the round trip is:
\[ \boxed{-5\sqrt{3} \, \text{J}} \]