Question

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

• A. 10 m/s
• B. 4 m/s
• C. 6 m/s
• D. 8 m/s

Hint:

Use the work-energy theorem to solve this problem. First, calculate the work done by the retarding force over the given distance by integrating the force with respect to displacement. Then, equate this work done to the change in kinetic energy of the block to find the final speed.

Answer 1

The Correct Option is D

The work done by the retarding force on the block as it moves through the rough region is given by: \[ W = \int_{x_i}^{x_f} F_r dx = \int_{0.1}^{1.9} (-kx) dx \] Substituting \( k = 10 \) N/m: \[ W = \int_{0.1}^{1.9} (-10x) dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9} \] \[ W = -5 \left[ (1.9)^2 - (0.1)^2 \right] = -5 [3.61 - 0.01] = -5 [3.60] = -18 \, \text{J} \] The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy: \[ W_{net} = \Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \] Given mass \( m = 1 \) kg and initial speed \( v_i = 10 \) m/s. The work done by the retarding force is the net work done on the block. \[ -18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10)^2 \] \[ -18 = \frac{1}{2} v_f^2 - \frac{1}{2} (100) \] \[ -18 = \frac{1}{2} v_f^2 - 50 \] \[ \frac{1}{2} v_f^2 = 50 - 18 = 32 \] \[ v_f^2 = 64 \] \[ v_f = \sqrt{64} = 8 \, \text{m/s} \] The final speed of the block as it crosses the rough region is 8 m/s.

Answer 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.