A particle of mass 2 kg is moving along x axis.its position at any time t is given as x=t°2.the work done by the force acting on it in time interval t=0 to t=5 s is

To determine the work done by the force acting on a particle moving along the x-axis, we first need to analyze the motion of the particle based on the given position function. The position of the particle is described by the equation \( x(t) = t^2 \), where \( t \) is the time in seconds. Let's break this down step by step.

Understanding the Motion

The position function \( x(t) = t^2 \) indicates that the particle's position changes with the square of time. To find the velocity, we differentiate the position function with respect to time:

Calculating Velocity

The velocity \( v(t) \) is the first derivative of the position function:

• \( v(t) = \frac{dx}{dt} = \frac{d(t^2)}{dt} = 2t \)

This tells us that the velocity of the particle increases linearly with time.

Finding Acceleration

Next, we calculate the acceleration \( a(t) \) by differentiating the velocity function:

• \( a(t) = \frac{dv}{dt} = \frac{d(2t)}{dt} = 2 \)

The acceleration is constant at \( 2 \, \text{m/s}^2 \), indicating that the particle is experiencing a uniform force.

Calculating Force

Using Newton's second law, \( F = ma \), we can find the force acting on the particle:

• Mass \( m = 2 \, \text{kg} \)
• Acceleration \( a = 2 \, \text{m/s}^2 \)
• Therefore, \( F = 2 \, \text{kg} \times 2 \, \text{m/s}^2 = 4 \, \text{N} \)

Calculating Work Done

The work done by a force is given by the formula:

• \( W = F \cdot d \)

Where \( d \) is the displacement of the particle over the time interval from \( t = 0 \) to \( t = 5 \) seconds. First, we need to find the displacement:

Finding Displacement

We can calculate the displacement by evaluating the position function at the endpoints of the time interval:

• At \( t = 0 \): \( x(0) = 0^2 = 0 \)
• At \( t = 5 \): \( x(5) = 5^2 = 25 \)

Thus, the total displacement \( d \) is:

• \( d = x(5) - x(0) = 25 - 0 = 25 \, \text{m} \)

Final Calculation of Work Done

Now we can substitute the values of force and displacement into the work formula:

• \( W = F \cdot d = 4 \, \text{N} \times 25 \, \text{m} = 100 \, \text{J} \)

In summary, the work done by the force acting on the particle during the time interval from \( t = 0 \) to \( t = 5 \) seconds is \( 100 \, \text{J} \).