A body of mass 5 kg moving alonga straight line is accelerated from `4ms^(-1)` to `8ms^(-1)` with the application of a froce of 10 N in the same direction. Then

Options

A. Work done by the fores is 120 J
B. Displacement of the body is 12m
C. Time for the change in velocity is 2s
D. All the above

Answer

To solve the problem step by step, we will follow the outlined process to find the displacement, work done, and time taken for the body.

### Step 1: Calculate the acceleration
We know the force applied and the mass of the body. We can use Newton's second law to find the acceleration.

\[ \text{Force} = \text{mass} \times \text{acceleration} \]

Given:
- Force (F) = 10 N
- Mass (m) = 5 kg

Using the formula:

\[ a = \frac{F}{m} = \frac{10 \, \text{N}}{5 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Step 2: Calculate the displacement using the kinematic equation
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity = 8 m/s
- \( u \) = initial velocity = 4 m/s
- \( a \) = acceleration = 2 m/s²
- \( s \) = displacement (what we need to find)

Substituting the values into the equation:

\[ 8^2 = 4^2 + 2 \cdot 2 \cdot s \]

Calculating:

\[ 64 = 16 + 4s \]

Rearranging to solve for \( s \):

\[ 64 - 16 = 4s \]
\[ 48 = 4s \]
\[ s = \frac{48}{4} = 12 \, \text{m} \]

### Step 3: Calculate the work done
Work done (W) is given by the formula:

\[ W = \text{Force} \times \text{Displacement} \]

Substituting the values:

\[ W = 10 \, \text{N} \times 12 \, \text{m} = 120 \, \text{J} \]

### Step 4: Calculate the time taken for the change in velocity
We can use the first equation of motion to find the time taken:

\[ v = u + at \]

Rearranging to solve for \( t \):

\[ t = \frac{v - u}{a} \]

Substituting the values:

\[ t = \frac{8 \, \text{m/s} - 4 \, \text{m/s}}{2 \, \text{m/s}^2} = \frac{4}{2} = 2 \, \text{s} \]

### Summary of Results
- Displacement \( s = 12 \, \text{m} \)
- Work done \( W = 120 \, \text{J} \)
- Time taken \( t = 2 \, \text{s} \)