JEE Advanced (2010)

LEVELJEE Advanced

Animated Solution for Physics - Work, Energy, and Power: A block of mass $2 kg$ is free to move along the $x$ -axis. It is at rest and from $t = 0$ onwards it is subjected to a time-dependent force $F (t)$ in the $x$ -direction. The force $F (t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 s$ is

Q: Why do we subtract the second area instead of adding it?

The force in the second region (from $t=3$ to $t=4.5\text{ s}$) is negative, meaning it acts in the opposite direction to the initial motion. This negative force decelerates the block, reducing its momentum. Therefore, the area is mathematically negative and must be subtracted.

Q: How did we find the force at $t=4.5\text{ s}$?

The $F-t$ graph is a single straight line, meaning its slope is constant. We calculated the slope using the points $(0, 4)$ and $(3, 0)$, which gave $-\frac{4}{3}$. We then used this slope to find the $y$-coordinate (Force) when the $x$-coordinate (time) is $4.5$.

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

If the block had an initial velocity $u$, it would have an initial momentum $p_i = mu$. The area under the graph would still give the *change* in momentum $\Delta p$. You would then find the final momentum using $p_f = p_i + \Delta p$, and finally calculate the kinetic energy using $K = \frac{p_f^2}{2m}$.

Select Answer:

Visualized Solution

Understanding the $F - t$ Graph

The problem provides a Force-time ( $F - t$ ) graph for a block of mass $m = 2 kg$ .
The block starts from rest, so initial velocity $u = 0$ and initial momentum $p_{i} = 0$ .
We need to find the kinetic energy at $t = 4.5 s$ .

Impulse-Momentum Theorem

According to Newton's Second Law, $F = \frac{d p}{d t}$ , which gives $d p = F d t$ .
Integrating both sides, $Δ p = \int F d t$ .
The integral $\int F d t$ represents the area under the $F - t$ graph.
Therefore, Change in Momentum = Area under $F - t$ graph.

Identifying the Areas

The total area consists of a positive area above the time axis ( $A_{1}$ ) and a negative area below it ( $A_{2}$ ).
$A_{1}$ is a triangle from $t = 0$ to $t = 3 s$ .
$A_{2}$ is a triangle from $t = 3 s$ to $t = 4.5 s$ .
To find $A_{2}$ , we need the value of force at $t = 4.5 s$ .

Force at $t = 4.5 s$

The slope of the $F - t$ graph is constant: $m = \frac{0 - 4}{3 - 0} = - \frac{4}{3} N/s$ .
Equation of the line: $F (t) = - \frac{4}{3} t + 4$ .
At $t = 4.5 s$ : $F (4.5) = - \frac{4}{3} (4.5) + 4 = - 6 + 4 = - 2 N$ .

Calculating Area $A_{1}$ and $A_{2}$

Area $A_{1} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 4 = 6 Ns$ .
Area $A_{2} = \frac{1}{2} \times base \times height = \frac{1}{2} \times (4.5 - 3) \times (- 2) = \frac{1}{2} \times 1.5 \times (- 2) = - 1.5 Ns$ .

Net Change in Momentum

Total Area $A = A_{1} + A_{2} = 6 - 1.5 = 4.5 Ns$ .
Since initial momentum $p_{i} = 0$ , the final momentum $p_{f} = Δ p = 4.5 kg m/s$ .

Relating Momentum to Kinetic Energy

Kinetic Energy $K$ is related to momentum $p$ by the formula: $K = \frac{p ^{2}}{2 m}$ .
We know $p = 4.5 kg m/s$ and $m = 2 kg$ .

Final Kinetic Energy

Substitute the values: $K = \frac{( 4.5 ) ^{2}}{2 \times 2}$ .
$K = \frac{20.25}{4}$ .
$K = 5.0625 J \approx 5.06 J$ .

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The Sigma Insight: Kinetic Energy, Potential Energy and Power

Solution Diagram

Analyzing the Setup

The problem presents a block of mass

m = 2 kg

initially at rest. A time-varying force

F (t)

acts on it, and we are given its graph. Our ultimate goal is to find the kinetic energy of the block at

t = 4.5 s

To bridge the gap between a force-time graph and kinetic energy, we must rely on the Impulse-Momentum Theorem.

The Master Equation

According to Newton's Second Law, force is the rate of change of momentum:

F = \frac{d p}{d t}

By integrating both sides with respect to time, we get the impulse, which equals the change in momentum:

Δ p = \int F d t

Graphically, the integral

\int F d t

is exactly the area under the

F - t

graph. Since the block starts from rest, its initial momentum is zero, meaning the final momentum

p

is simply equal to this area.

Decoding the Graph

The graph consists of two triangular regions: a positive area

A_{1}

above the time axis (from

t = 0

t = 3 s

) and a negative area

A_{2}

below the time axis (from

t = 3 s

t = 4.5 s

Before calculating

A_{2}

, we need the height of the second triangle, which corresponds to the force at

t = 4.5 s

. The graph is a straight line, so its slope is constant:

m = \frac{0 - 4}{3 - 0} = - \frac{4}{3} N/s

Using the point-slope form, the force at

t = 4.5 s

is:

F (4.5) = - \frac{4}{3} (4.5 - 3) + 0 = - \frac{4}{3} (1.5) = - 2 N

Calculating the Impulse

Now we can compute the areas of the two triangles.

The positive area

A_{1}

is:

A_{1} = \frac{1}{2} \times 3 \times 4 = 6 Ns

The negative area

A_{2}

is:

A_{2} = \frac{1}{2} \times 1.5 \times (- 2) = - 1.5 Ns

The net area, and thus the final momentum

p

, is the sum of these areas:

p = A_{1} + A_{2} = 6 - 1.5 = 4.5 kg m/s

Final Calculation

With the momentum in hand, we can find the kinetic energy using the classic relation:

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

Substituting the values

p = 4.5 kg m/s

and

m = 2 kg

K = \frac{( 4.5 ) ^{2}}{2 \times 2} = \frac{20.25}{4} = 5.0625 J

Rounding to two decimal places, the kinetic energy is $5.06 J$ , which perfectly matches option (c).

Common Traps & FAQs

Why do we subtract the second area instead of adding it?

The force in the second region (from

t = 3

t = 4.5 s

) is negative, meaning it acts in the opposite direction to the initial motion. This negative force decelerates the block, reducing its momentum. Therefore, the area is mathematically negative and must be subtracted.

How did we find the force at

t = 4.5 s

The

F - t

graph is a single straight line, meaning its slope is constant. We calculated the slope using the points

(0, 4)

and

(3, 0)

, which gave

- \frac{4}{3}

. We then used this slope to find the

y

-coordinate (Force) when the

x

-coordinate (time) is

4.5

Can we use

v = u + a t

to find the velocity?

No, because the acceleration is not constant! The force is changing with time, which means the acceleration is also changing. The equations of kinematics (

v = u + a t

, etc.) are strictly for constant acceleration. We must use calculus or graphical methods (like area under the curve) here.

Why use

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

instead of

K = \frac{1}{2} m v^{2}

Both formulas are perfectly valid and will give the exact same answer! However, since the area under the

F - t

graph directly gives us momentum

p

, using

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

saves us the extra step of calculating velocity

v = \frac{p}{m}

. It's a faster and more elegant route.

What if the block didn't start from rest?

If the block had an initial velocity

u

, it would have an initial momentum

p_{i} = m u

. The area under the graph would still give the *change* in momentum

Δ p

. You would then find the final momentum using

p_{f} = p_{i} + Δ p

, and finally calculate the kinetic energy using

K = \frac{p _{f}^{2}}{2 m}

Select Answer:

Visualized Solution

Understanding the $F - t$ Graph

Impulse-Momentum Theorem

Identifying the Areas

Force at $t = 4.5 s$

Calculating Area $A_{1}$ and $A_{2}$

Net Change in Momentum

Relating Momentum to Kinetic Energy

Final Kinetic Energy

The Sigma Insight: Kinetic Energy, Potential Energy and Power

Analyzing the Setup

The Master Equation

Decoding the Graph

Calculating the Impulse

Final Calculation

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Select Answer:

Visualized Solution

Understanding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F−t Graph

Impulse-Momentum Theorem

Identifying the Areas

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Calculating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

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Relating Momentum to Kinetic Energy

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