JEE Physics Solution – Question 16

Solution Transcription

1. Define Variables from the Graph

Let one vertical division on the y-axis (velocity) be denoted by $c$.
Let the force of air resistance be denoted by a general function $F(v)$.

2. Formulate the Equation of Motion at t = 5.0 s

From the handwritten solution, the acceleration $a$ at $t=5.0 \text{ s}$ (derived from the slope of the tangent AB) is given as: $$a = \frac{3c}{4}$$ Applying Newton’s Second Law ($F_{net} = ma$) at this instant: $$mg – F_{air} = ma$$

Substituting the expression for acceleration:

$$\frac{mg – F(v)}{m} = \frac{3c}{4}$$ $$\frac{F(v)}{m} = g – \frac{3c}{4}$$

3. Determine the Scale Constant ‘c’

We use the initial part of the graph ($t=0$ to $t=1 \text{ s}$) to find the value of $c$. At $t=0$, velocity is zero, so air resistance is zero. The initial acceleration is equal to gravity ($g$).

From the graph, at $t=1 \text{ s}$, the velocity is 2 divisions ($2c$). Approximating the initial acceleration:

$$a_{initial} \approx \frac{\Delta v}{\Delta t} = \frac{2c – 0}{1 – 0} = 2c$$ $$\therefore g = 2c \implies c = \frac{g}{2}$$

4. Calculate the Force of Air Resistance

Now, substitute the value of $c = g/2$ back into the equation from Step 2 to find the force.

$$\frac{F(v)}{m} = g – \frac{3}{4}\left(\frac{g}{2}\right) = g – \frac{3g}{8}$$ $$\frac{F(v)}{m} = \frac{5g}{8}$$ $$\therefore F_{air} = \frac{5mg}{8}$$

5. Calculate the Power

Power is calculated as the product of force and velocity ($P = F \cdot v$). Following the handwritten solution, we use the terminal velocity $v = 8c$ for this calculation.

$$P = F_{air} \times (8c)$$ $$P = \left( \frac{5mg}{8} \right) \times 8c = 5mgc$$

Substituting the given values ($m=60 \text{ kg}$, $g=10 \text{ m/s}^2$) and $c = g/2 = 5$:

$$P = 5 \times (60) \times (10) \times (5)$$ $$P = 15000 \text{ W}$$

Power = 15 kW