A block of mass 1 kg, moving along x with speed \( v_i = 10 \) m/s enters a rough region ranging from \( x = 0.1 \) m to \( x = 1.9 \) m. The retarding force acting on the block in this range is \( F_r = -kx \) N, with \( k = 10 \) N/m. Then the final speed of the block as it crosses this rough region is
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- 10 m/s
- 4 m/s
- 6 m/s
- 8 m/s
The Correct Option is D
Approach Solution - 1
Approach Solution -2
To find the final speed of the block after crossing the rough region, we need to use the work-energy principle. The work done by the retarding force is equal to the change in kinetic energy of the block.
If the initial kinetic energy is \( KE_i = \frac{1}{2} m v_i^2 \) and the final kinetic energy is \( KE_f = \frac{1}{2} m v_f^2 \), the work done by the retarding force \( W = \int F_r \, dx \) over the range from \( x = 0.1 \) m to \( x = 1.9 \) m, can be calculated.
The retarding force is given by \( F_r = -kx \). Therefore, the work done by this force over the distance is:
\(W = \int_{0.1}^{1.9} -kx \, dx = -k \left[\frac{x^2}{2}\right]_{0.1}^{1.9}\)
\(= -10 \left[\frac{1.9^2}{2} - \frac{0.1^2}{2}\right]\)
\(= -10 \left[\frac{3.61}{2} - \frac{0.01}{2}\right]\)
\(= -10 \left[1.805 - 0.005\right]\)
\(= -10 \times 1.8 = -18 \, \text{J}\)
Now, according to the work-energy principle:
\(\Delta KE = KE_f - KE_i = W\)
Substitute the values:
\(\frac{1}{2} m v_f^2 - \frac{1}{2} \times 1 \times 10^2 = -18\)
\(\frac{1}{2} \times 1 \times v_f^2 - 50 = -18\)
\(\frac{1}{2} v_f^2 = 32\)
\(v_f^2 = 64\)
\(v_f = \sqrt{64} = 8 \, \text{m/s}\)
Therefore, the final speed of the block as it crosses the rough region is 8 m/s.
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