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Step-by-Step Detailed Solution
Step 1: Identify the known quantities
Mass of the block, $m = 2 \,\text{kg}$
Initial speed of the block, $u = 4 \,\text{m/s}$
Range of rough surface: $x$ from $0.5 \,\text{m}$ to $1.5 \,\text{m}$
Retarding force: $F(x) = -kx$ with $k = 12 \,\text{N/m}$
We need to find the final speed $v$ of the block just after it crosses $x = 1.5 \,\text{m}$
Step 2: Understand the relation between force, mass, and acceleration
From Newton's second law, $F = ma$. For a varying force $F(x) = m \frac{dv}{dt}$. However, when velocity $v$ is involved and force depends on $x$, we often use the form:
$m \, v \frac{dv}{dx} = F(x)\,.$
Here, $F(x) = -12x$.
Step 3: Set up the differential equation
Using $m \, v \frac{dv}{dx} = -12x$, we substitute $m = 2$ kg:
$2 \, v \frac{dv}{dx} = -12x \quad \Longrightarrow \quad v \frac{dv}{dx} = -6x\,.$
Step 4: Apply integration limits and solve
We separate variables and integrate from the region where $x$ goes from $0.5\,\text{m}$ to $1.5\,\text{m}$, while the speed changes from $4\,\text{m/s}$ to $v\,\text{m/s}$.
Rewriting:
$v \, dv = -6x \, dx \,.$
Integrate with respect to $v$ and $x$:
$\int_{4}^{v} v \, dv = -6 \int_{0.5}^{1.5} x \, dx\,.$
Perform the integrals:
\[
\int_{4}^{v} v \, dv = \left[\frac{v^2}{2}\right]_{4}^{v} = \frac{v^2}{2} - \frac{16}{2} = \frac{v^2 - 16}{2},
\]
\[
\int_{0.5}^{1.5} x \, dx = \left[\frac{x^2}{2}\right]_{0.5}^{1.5} = \frac{1.5^2 - 0.5^2}{2} = \frac{2.25 - 0.25}{2} = \frac{2}{2} = 1.
\]
Hence,
\[
\frac{v^2 - 16}{2} = -6 \times 1 = -6.
\]
Step 5: Solve for the final speed $v$
From
\[
\frac{v^2 - 16}{2} = -6,
\]
we get
\[
v^2 - 16 = -12 \quad \Rightarrow \quad v^2 = 4 \quad \Rightarrow \quad v = 2\,\text{m/s}
\]
(taking the positive root because speed cannot be negative).
Step 6: State the final answer
The speed of the block as it just crosses the rough surface is $2.0 \,\text{m/s}$.
