A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the

Q: Why can't we just use $W = F \times d$?

The formula $W = F \times d$ is only valid for a *constant* force. Here, the force $F = 6t$ changes with time. To use the force-displacement method, you would need to find the displacement as a function of time and integrate $W = \int F dx$, which is a bit longer!

Q: Why did we calculate momentum instead of velocity?

Both methods are perfectly valid! Finding velocity ($v = \int a dt$) and then using $K = \frac{1}{2}mv^2$ gives the exact same result. The momentum approach ($K = \frac{p^2}{2m}$) is just a neat alternative that directly uses Newton's second law in its original form.

Q: What if the particle didn't start from rest?

If it had an initial velocity, it would have an initial momentum $p_i$. You would calculate the final momentum as $p_f = p_i + \int F dt$, and then the change in kinetic energy would be $\frac{p_f^2}{2m} - \frac{p_i^2}{2m}$.

Q: Is the Work-Energy theorem always applicable?

Yes! The Work-Energy theorem ($W_{\text{net}} = \Delta K$) is a universal principle in mechanics. It applies to constant forces, variable forces, conservative forces, and non-conservative forces alike.

Q: Why is the integration limit from 0 to 1?

The question specifically asks for the work done during the *first* 1 second. This means the time interval starts at $t = 0$ and ends at $t = 1\text{ s}$.

Question

A time dependent force $F = 6t$ acts on a particle of mass $1	ext{ kg}$. If the particle starts from rest, the work done by the force during the first $1	ext{ s}$ will be

Options:

(A) 22 J

(B) 9 J

(C) 18 J

(D) 4.5 J

Sigma Percentile · Accepted Answer

### The Setup: A Time-Dependent Force

Imagine a block of mass $m = 1	ext{ kg}$ resting peacefully on a frictionless horizontal surface. Suddenly, a force begins to push it. But this isn't just any constant force; it's a time-dependent force given by $F = 6t$. This means the force starts at zero and grows stronger with every passing second. Our mission is to find the total work done by this force during the very first second of its journey.

### Newton's Second Law: The Momentum Connection

When dealing with forces that change with time, Newton's Second Law in its original momentum form is an incredibly powerful tool. We know that force is the rate of change of momentum:

$$F = \frac{dp}{dt}$$

By rearranging this, we can relate a tiny change in momentum $dp$ to a tiny time interval $dt$:

$$dp = F dt$$

To find the total momentum acquired by the block after $1	ext{ s}$, we simply integrate both sides. Since the block starts from rest, its initial momentum at $t = 0$ is zero.

$$\int_0^p dp = \int_0^1 6t dt$$

Evaluating the integral is straightforward:

$$p = \left[ \frac{6t^2}{2} ight]_0^1 = 3(1)^2 - 0 = 3	ext{ kg m/s}$$

So, at exactly $t = 1	ext{ s}$, the block has built up a momentum of $3	ext{ kg m/s}$.

### The Kinetic Energy Link

Now that we have the momentum, how do we find the work done? We need a bridge between momentum and energy. That bridge is the kinetic energy formula expressed in terms of momentum:

$$K = \frac{p^2}{2m}$$

Let's calculate the final kinetic energy of our block at $t = 1	ext{ s}$:

$$K_f = \frac{3^2}{2(1)} = \frac{9}{2} = 4.5	ext{ J}$$

Since the block started from rest, its initial kinetic energy $K_i$ was $0$. Therefore, the total change in kinetic energy $\Delta K$ is simply $4.5	ext{ J}$.

### The Grand Finale: Work-Energy Theorem

We are at the final step! The Work-Energy Theorem is one of the most elegant principles in physics. It states that the net work done on an object is exactly equal to its change in kinetic energy:

$$W = \Delta K$$

Since we just calculated $\Delta K$ to be $4.5	ext{ J}$, the work done by our time-dependent force is exactly $4.5	ext{ J}$.

**Final Answer:** The work done by the force during the first $1	ext{ s}$ is **$4.5	ext{ J}$**.

Animated Solution for Physics - Work, Energy, and Power: A time dependent force $F = 6 t$ acts on a particle of mass $1 kg$ . If the particle starts from rest, the work done by the force during the first $1 s$ will be

Select Answer:

Visualized Solution

Visualizing the Setup

Newton's Second Law

Integrating for Momentum

Calculating Final Momentum

Kinetic Energy Relation

Calculating Change in KE

Work-Energy Theorem

Final Answer

The Sigma Insight: Work Done by Forces

The Setup

Newton's Second Law

The Kinetic Energy Link

The Grand Finale

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