Animated Solution for Physics - Laws of Motion: A ship of mass $3\times 10^7$ kg initially at rest, is pulled by a force of $5\times 10^4$ N through a distance of $3$ m. Assuming that the resistance due to water is negligible, the speed of the ship is

Moving a Mountain on Water

Imagine standing at the docks, looking up at a colossal steel behemoth resting peacefully on the water. This is no ordinary boat; it is a massive ship with a mass of $3\times 10^7\text{ kg}$. To put that into perspective, that is thirty million kilograms of solid mass!

By its very nature, this ship possesses an enormous amount of inertia. Inertia is the stubbornness of an object to change its state of motion. Because it is initially at rest, it desperately wants to stay at rest.

However, a tugboat arrives and begins to pull this giant with a force of $5\times 10^4\text{ N}$. Fifty thousand Newtons might sound like a lot of force—and it is, if you were pushing a car—but against thirty million kilograms, it is a David versus Goliath scenario.

Our mission is to determine exactly how fast this ship will be moving after it has been dragged through a relatively short distance of $3\text{ m}$. We are also given a crucial simplifying assumption: the resistance due to the water is negligible. This means we don't have to worry about drag forces stealing our energy; every single Newton of that pulling force goes directly into accelerating the ship.

Newton's Second Law

To find the final speed of the ship, we must first understand how its motion is changing. This brings us to the cornerstone of classical mechanics: Newton's Second Law of Motion.

Sir Isaac Newton taught us that the net force acting on an object is directly proportional to the acceleration it experiences, and the proportionality constant is the object's mass. Mathematically, this is expressed in the legendary equation:

In our scenario, we know the force $F$ and the mass $m$. What we desperately need is the acceleration $a$. Acceleration is the rate at which the ship's velocity increases. By rearranging Newton's Second Law, we can isolate acceleration:

This equation tells a beautiful story. It says that the acceleration is directly proportional to the force (pull harder, and it speeds up faster) but inversely proportional to the mass (the heavier the object, the more sluggish its response).

Let us substitute the values given in the problem into our rearranged equation:

Now, we must handle the scientific notation carefully. We can separate the coefficients from the powers of ten:

Subtracting the exponents gives us:

This is a remarkably tiny acceleration! It is just a fraction of a millimeter per second squared. But this makes perfect physical sense. When you apply a moderate force to an incredibly massive object, the resulting acceleration will be minuscule. The ship is speeding up, but it is doing so at a glacial pace.

Kinematics in Action

Now that we have unlocked the acceleration, we hold the key to the ship's future. We need to find its final velocity $v$ after it has traveled a displacement $s=3\text{ m}$.

Let us take an inventory of our known kinematic variables: 1. Initial velocity, $u=0\text{ m/s}$ (since the ship is initially at rest). 2. Acceleration, $a=\frac{5}{3}\times 10^{-3}{\text{ m/s}}^2$. 3. Displacement, $s=3\text{ m}$.

We are looking for the final velocity $v$. Notice that we do not know the time $t$ it took for this motion to occur, and we are not asked to find it.

This is where we must select the right tool from our kinematic arsenal. The equations of motion are our best friends here. The first equation ($v=u+at$) and the second equation ($s=ut+\frac{1}{2}a{t}^2$) both require time.

However, the third equation of motion elegantly bypasses time altogether, linking velocity directly to displacement:

This equation is perfect for our situation. It is a direct bridge from what we know to what we want to find.

Calculating the Speed

Let us substitute our known values into the third equation of motion. We must do this carefully to avoid any algebraic missteps.

The zero squared simply vanishes, leaving us with the core multiplication. Look closely at the structure of the expression on the right side. We have a $3$ in the denominator of our acceleration term, and we are multiplying the entire expression by a displacement of $3$.

This is a moment of mathematical elegance. The $3$ in the numerator and the $3$ in the denominator cancel each other out perfectly!

Now, we simply multiply the remaining coefficients. Two times five is ten.

Using the laws of exponents, we know that $10$ is the same as $10^1$. When we multiply numbers with the same base, we add their exponents: $1+(-3)=-2$.

We are at the final threshold. We have the square of the velocity, but we need the velocity itself. To find $v$, we must take the square root of both sides of the equation.

Taking the square root of a power of ten simply means dividing the exponent by two. Therefore, the square root of $10^{-2}$ is $10^{-1}$.

Finally, we can convert this scientific notation back into a standard decimal format. Ten to the power of negative one is exactly one-tenth.

And there we have it! After being pulled for three meters, the colossal ship is moving at a speed of $0.1\text{ m/s}$. It is a slow, majestic crawl, perfectly consistent with the immense inertia of the vessel. The correct option is (c).