Animated Solution for Physics - Work, Energy, and Power: 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?

Imagine you are tracking a $2\text{ kg}$ block sliding along a frictionless track. You don't have a force sensor, nor do you have a speedometer. All you have is a mathematical blueprint of its journey: its position as a function of time, given by $x(t)=3t^2+5$.

From this simple equation, we are tasked with finding the total work done by the invisible force pushing this block over the first $5$ seconds. While it might seem like we lack information, the beauty of classical mechanics is that kinematics and dynamics are deeply intertwined. Let's unravel this step-by-step.

When asked to find the work done, your first instinct might be to reach for the classic formula $W=\int Fdx$. While perfectly valid, it requires us to find the force and the displacement separately.

Instead, let's use the ultimate shortcut in physics: the Work-Energy Theorem. This theorem states that the net work done on an object is exactly equal to its change in kinetic energy:

To use this elegant tool, we only need two things: the initial velocity ($v_i$) and the final velocity ($v_f$).

We know the position function $x(t)=3t^2+5$. Velocity is simply the rate at which position changes, which means we need to take the time derivative of $x(t)$:

Now, let's find our boundary velocities. At the very beginning of our observation ($t=0\text{ s}$):

The block starts from rest. Now, let's look at the end of our time window ($t=5\text{ s}$):

The block has accelerated to a brisk $30\text{ m/s}$.

We have all the pieces of our puzzle. Let's plug them back into the Work-Energy Theorem. The mass $m$ is $2\text{ kg}$:

Notice how the mass of $2$ perfectly cancels out the $\frac{1}{2}$ in the kinetic energy formula. We are left with a simple square:

And there we have it! The total work done by the force is exactly $900\text{ Joules}$.

Could we have solved this using Newton's Laws directly? Absolutely. If we differentiate the velocity $v(t)=6t$, we get a constant acceleration $a=6{\text{ m/s}}^2$.

Using Newton's Second Law, the constant force is $F=ma=2\times 6=12\text{ N}$.

The displacement over $5$ seconds is $x(5)-x(0)=80-5=75\text{ m}$.

Multiplying the constant force by the displacement gives $W=F\cdot s=12\times 75=900\text{ J}$. Physics is beautifully consistent, no matter which path you choose to walk!