A body of mass 2 kg is driven by an engine delivering a constant power of 1 J/s. The body starts fro

Q: Why can't we use the standard equations of motion like $s = ut + \frac{1}{2}at^2$?

Those equations are strictly for constant acceleration. Here, constant power means the force (and thus acceleration) is changing as the velocity changes ($P = Fv \Rightarrow F = P/v$). Since acceleration is not constant, those equations are invalid.

Q: How did we get $P = m v^2 \frac{dv}{ds}$?

We start with $P = Fv$. By Newton's second law, $F = ma$. Acceleration $a$ can be written as $\frac{dv}{dt}$ or, using the chain rule, $v\frac{dv}{ds}$. Substituting this gives $P = m(v\frac{dv}{ds})v = m v^2 \frac{dv}{ds}$.

Q: Is the work done by the engine the only work done on the body?

Yes, the problem states the body moves in a straight line and doesn't mention any resistive forces like friction. So, the net work done is solely by the engine.

Q: Why did we integrate from $v=0$ to $v=3$?

We found the velocity at $t=9\text{ s}$ is $3\text{ m/s}$. The body started from rest ($v=0$). To find the total distance covered in this interval, we integrate the displacement element $ds$ over the corresponding velocity limits.

Q: Could we have integrated with respect to time instead of velocity?

Absolutely! From $P = \frac{1}{2}mv^2/t$, you get $v = \sqrt{2Pt/m}$. Since $v = ds/dt$, you can write $ds = \sqrt{2Pt/m} dt$ and integrate from $t=0$ to $t=9$. Both methods yield the exact same result.

Question

A body of mass $2	ext{ kg}$ is driven by an engine delivering a constant power of $1	ext{ J/s}$. The body starts from rest and moves in a straight line. After $9	ext{ s}$, the body has moved a distance (in m) ......... .

Sigma Percentile · Accepted Answer

Imagine a $2	ext{ kg}$ block resting peacefully on a perfectly smooth surface. Suddenly, an engine kicks in, delivering a constant power of $1	ext{ W}$. The block begins to accelerate, but because the power is constant, the force pushing it must decrease as it speeds up. This isn't your standard constant-acceleration kinematics problem. We are dealing with a dynamic, changing force. Our mission? To find out exactly how far this block travels in $9	ext{ s}$.

### The Work-Energy Route

The most elegant way to tackle this is by looking at the energy of the system. We know that the engine is doing work on the block, and since the surface is smooth, all this work goes directly into increasing the block's kinetic energy.

The Work-Energy Theorem states:
$$W = \Delta K$$

Because the power $P$ is constant, the total work done over a time $t$ is simply $P 	imes t$. Therefore:
$$P \cdot t = \frac{1}{2}mv^2 - 0$$

Let's plug in our known values: $P = 1	ext{ W}$, $t = 9	ext{ s}$, and $m = 2	ext{ kg}$.
$$1 \cdot 9 = \frac{1}{2}(2)v^2$$
$$9 = v^2 \Rightarrow v = 3	ext{ m/s}$$

So, at exactly $t = 9	ext{ s}$, the block is cruising at $3	ext{ m/s}$. But we need the distance, not just the final velocity.

### Linking Power, Velocity, and Position

To find the distance, we need to dig into the fundamental definition of power. Power is the rate at which work is done, which can also be expressed as force times velocity:
$$P = F \cdot v$$

By Newton's Second Law, force is mass times acceleration ($F = ma$). And acceleration can be written using the chain rule as $v \frac{dv}{ds}$. Substituting this in, we get a beautiful differential equation:
$$P = m \left(v \frac{dv}{ds}ight) v = m v^2 \frac{dv}{ds}$$

Now, let's rearrange this to isolate the small displacement element, $ds$:
$$ds = \frac{m}{P} v^2 dv$$

Substituting $m = 2	ext{ kg}$ and $P = 1	ext{ W}$, we get:
$$ds = 2 v^2 dv$$

### The Final Integration

We are now ready to integrate. The block starts from rest ($v = 0$) at position $s = 0$, and reaches a velocity of $v = 3	ext{ m/s}$ at our unknown final position $s$.
$$\int_0^s ds = \int_0^3 2v^2 dv$$

Evaluating the integral on the right side:
$$s = 2 \left[ \frac{v^3}{3} ight]_0^3$$
$$s = \frac{2}{3} (3^3 - 0^3) = \frac{2}{3} (27)$$
$$s = 18	ext{ m}$$

The block travels exactly $18	ext{ m}$ in those $9	ext{ s}$.

### Pro-Tip: The Time-Space Route

Could we have solved this without finding the final velocity first? Absolutely! From our Work-Energy equation $Pt = \frac{1}{2}mv^2$, we can express velocity directly as a function of time:
$$v(t) = \sqrt{\frac{2P}{m}} t^{1/2}$$

Since velocity is the derivative of position ($v = \frac{ds}{dt}$), we can integrate this expression with respect to time to find the total distance:
$$s = \int_0^t \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \frac{2}{3} t^{3/2}$$

Plugging in $P=1$, $m=2$, and $t=9$:
$$s = \sqrt{\frac{2(1)}{2}} \frac{2}{3} (9)^{3/2} = 1 \cdot \frac{2}{3} \cdot 27 = 18	ext{ m}$$

Both paths lead to the exact same elegant result. Physics is beautifully consistent!

Animated Solution for Physics - Work, Energy, and Power: A body of mass $2 kg$ is driven by an engine delivering a constant power of $1 J/s$ . The body starts from rest and moves in a straight line. After $9 s$ , the body has moved a distance (in m) ......... .

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Work-Energy Theorem}

\text{Velocity at } t = 9\text{ s}

\text{Power in terms of Velocity}

\text{Setting up the Integral}

\text{Integrating for Distance}

\text{Final Calculation}

\text{Alternative Approach}

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Work-Energy Route

Linking Power, Velocity, and Position

The Final Integration

Pro-Tip

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