Question:
A block of mass \(2\,\text{kg}\), moving along the \(X\)-axis on a horizontal surface with a speed of \(4\,\text{ms}^{-1}\), enters a rough surface from \(x=0.5\,\text{m}\) to \(x=1.5\,\text{m}\). The retarding force on this rough surface is
\[
F=-12x\ \text{N}.
\]
The speed of the block as it just crosses the rough surface is
A block of mass \(2\,\text{kg}\), moving along the \(X\)-axis on a horizontal surface with a speed of \(4\,\text{ms}^{-1}\), enters a rough surface from \(x=0.5\,\text{m}\) to \(x=1.5\,\text{m}\). The retarding force on this rough surface is
\[
F=-12x\ \text{N}.
\]
The speed of the block as it just crosses the rough surface is
Show Hint
For a variable force, work is calculated using
\[
W=\int F\,dx.
\]
Then use the Work-Energy Theorem
\[
W=\Delta K
\]
to find the final speed directly without calculating acceleration.
Updated On: Jul 29, 2026
- Zero
- \(1.5\,\text{ms}^{-1}\)
- \(2\,\text{ms}^{-1}\)
- \(2.5\,\text{ms}^{-1}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Use the Work-Energy Theorem:
\[
W=\Delta K.
\]
The work done by the retarding force equals the change in kinetic energy.
Step 1: Calculate the initial kinetic energy. Given, \[ m=2\,\text{kg}, \qquad u=4\,\text{ms}^{-1}. \] Hence, \[ K_i = \frac12 mu^2 = \frac12(2)(4)^2 = 16\ \text{J}. \]
Step 2: Find the work done by the variable force. The force is \[ F=-12x. \] Therefore, \[ W = \int_{0.5}^{1.5}F\,dx = \int_{0.5}^{1.5}(-12x)\,dx. \] \[ = -12 \left[\frac{x^2}{2}\right]_{0.5}^{1.5}. \] \[ = -6 \left(1.5^2-0.5^2\right). \] \[ = -6(2.25-0.25). \] \[ = -12\ \text{J}. \]
Step 3: Apply the work-energy theorem. \[ W = K_f-K_i. \] \[ -12 = K_f-16. \] \[ K_f=4\ \text{J}. \]
Step 4: Find the final speed. \[ K_f = \frac12 mv^2. \] \[ 4 = \frac12(2)v^2. \] \[ v^2=4. \] \[ v=2\,\text{ms}^{-1}. \] Therefore, \[ \boxed{v=2\,\text{ms}^{-1}} \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Calculate the initial kinetic energy. Given, \[ m=2\,\text{kg}, \qquad u=4\,\text{ms}^{-1}. \] Hence, \[ K_i = \frac12 mu^2 = \frac12(2)(4)^2 = 16\ \text{J}. \]
Step 2: Find the work done by the variable force. The force is \[ F=-12x. \] Therefore, \[ W = \int_{0.5}^{1.5}F\,dx = \int_{0.5}^{1.5}(-12x)\,dx. \] \[ = -12 \left[\frac{x^2}{2}\right]_{0.5}^{1.5}. \] \[ = -6 \left(1.5^2-0.5^2\right). \] \[ = -6(2.25-0.25). \] \[ = -12\ \text{J}. \]
Step 3: Apply the work-energy theorem. \[ W = K_f-K_i. \] \[ -12 = K_f-16. \] \[ K_f=4\ \text{J}. \]
Step 4: Find the final speed. \[ K_f = \frac12 mv^2. \] \[ 4 = \frac12(2)v^2. \] \[ v^2=4. \] \[ v=2\,\text{ms}^{-1}. \] Therefore, \[ \boxed{v=2\,\text{ms}^{-1}} \] \[ \boxed{\text{Answer = (C)}} \]
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