The distance x moved by a body of mass 0.5 kg under the action of a force varies with time t as
`x(m)=3t^(2)+4t+5`
Here, t is expressed in second. What is the work done by the force in first 2 seconds?

Explanation

To solve the problem of calculating the work done by the force on a body of mass 0.5 kg that moves according to the equation \( x(t) = 3t^2 + 4t + 5 \) over the first 2 seconds, we can follow these steps:

### Step 1: Determine the displacement at \( t = 0 \) and \( t = 2 \) seconds.
We start by calculating the displacement at the initial time \( t = 0 \) and at \( t = 2 \) seconds using the given equation for displacement.

- At \( t = 0 \):
\[
x(0) = 3(0)^2 + 4(0) + 5 = 5 \, \text{m}
...