Question

A block of mass 2 kg moving on a horizontal surface with speed of 4 ms^–1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = –kx where k = 12 Nm^–1. The speed of the block as it just crosses the rough surface will be :

• A. Zero
• B. 1.5 ms^–1
• C. 2.0 ms^–1
• D. 2.5 ms^–1

Answer 1

The Correct Option is C

To solve this problem, we need to determine the speed of the block as it exits the rough surface. Let's break this down step-by-step.

1. Initial Setup and Given Data 

• Mass of the block, \(m = 2 \, \text{kg}\).
• Initial speed of the block, \(u = 4 \, \text{ms}^{-1}\).
• Range of the rough surface: from \(x = 0.5 \, \text{m}\) to \(x = 1.5 \, \text{m}\).
• Retarding force is given by \(F = -kx\) with \(k = 12 \, \text{Nm}^{-1}\).

2. Work Done by the Retarding Force

The work done by the force \(F = -kx\) as the block moves from \(x = 0.5 \, \text{m}\) to \(x = 1.5 \, \text{m}\) is calculated using the formula:

\(W = \int_{x_1}^{x_2} F \, dx = \int_{0.5}^{1.5} -kx \, dx\)

Substitute the value of \(k\):

\(W = \int_{0.5}^{1.5} -12x \, dx = \left[ -12 \cdot \frac{x^2}{2} \right]_{0.5}^{1.5}\)

Calculate the definite integral:

\(W = \left[ -6x^2 \right]_{0.5}^{1.5} = -6(1.5^2) + 6(0.5^2)\)

\(W = -6(2.25) + 6(0.25) = -13.5 + 1.5 = -12 \, \text{Joules}\)

3. Applying Energy Conservation

According to the work-energy principle, the work done by the retarding force changes the kinetic energy of the block. The change in kinetic energy is:

\(\Delta KE = KE_{\text{final}} - KE_{\text{initial}} = W\)

\(KE_{\text{initial}} = \frac{1}{2} m u^2 = \frac{1}{2} \times 2 \times 4^2 = 16 \, \text{Joules}\)

\(KE_{\text{final}} = KE_{\text{initial}} + W = 16 - 12 = 4 \, \text{Joules}\)

4. Calculating the Final Speed

The final kinetic energy is given by:

\(KE_{\text{final}} = \frac{1}{2} m v^2\)

Solve for \(v\):

\(\frac{1}{2} \times 2 \times v^2 = 4\)

\(v^2 = 4 \quad \Rightarrow \quad v = \sqrt{4} = 2 \, \text{ms}^{-1}\)

Conclusion

The speed of the block as it just crosses the rough surface is 2.0 ms^–1.

Answer 2

The correct answer is (C) : 2.0 ms^–1 

F = –12x

\[ mv \frac{dv}{dx} = -12 \]

\[ \int_{4}^v v \, dv = -6 \int_{0.5}^{1.5} x \, dx \]

(m = 2 kg)

\[ \frac{v^2 - 16}{2} = -6 \left[\frac{1.5^2 - 0.5^2}{2}\right] \]

\[ \frac{v^2 - 16}{2} = -6 \]

v = 2 m/sec