A force acts on a 2 kg object so that its position is given as a function of time as $$x = 3t^2 + 5$$. What is the work done by this force in first 5 seconds?

We are told that the position of the 2 kg object varies with time according to the law $$x = 3t^{2} + 5$$, where $$x$$ is in metres and $$t$$ in seconds. From this function we can find velocity, then acceleration, then force, and finally the work done.

First we differentiate the position function to obtain velocity, because by definition $$v = \dfrac{dx}{dt}$$.

So, $$v = \dfrac{d}{dt}\left(3t^{2} + 5\right) = 6t.$$

Next, we differentiate velocity with respect to time to obtain acceleration, since $$a = \dfrac{dv}{dt}$$.

Thus, $$a = \dfrac{d}{dt}\left(6t\right) = 6\; \text{m s}^{-2}.$$

Notice that the acceleration is a constant 6 m s−2. Because the mass is 2 kg, Newton’s second law $$F = ma$$ gives the magnitude of the force:

$$F = m a = 2 \times 6 = 12\;\text{N}.$$

We need the work done by this force in the time interval from $$t = 0$$ to $$t = 5\;\text{s}$$. A convenient route is to use the Work-Energy Theorem, which states:

“The net work done on a particle equals the change in its kinetic energy,” i.e. $$W = K_{\text{final}} - K_{\text{initial}}.$$

We already have the expression for velocity; let us evaluate it at the two instants.

At $$t = 0$$, $$v_{0} = 6(0) = 0\;\text{m s}^{-1}.$$

At $$t = 5\;\text{s}$$, $$v_{5} = 6(5) = 30\;\text{m s}^{-1}.$$

The kinetic energy formula is $$K = \tfrac{1}{2} m v^{2}.$$ Substituting the respective velocities:

Initial kinetic energy $$K_{0} = \tfrac{1}{2} \times 2 \times (0)^{2} = 0\;\text{J}.$$

Final kinetic energy $$K_{5} = \tfrac{1}{2} \times 2 \times (30)^{2} = 1 \times 900 = 900\;\text{J}.$$

Now we apply the Work-Energy Theorem:

$$W = K_{5} - K_{0} = 900 - 0 = 900\;\text{J}.$$

Hence, the correct answer is Option D.