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 rough region is.

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

Answer 1

Answer: (D)

8 m/s

Answer 2

Here's how to solve this problem using the work-energy theorem:

1. Calculate the work done by the retarding force:

   The work done by a variable force is given by the integral of the force over the distance:

   $W = \int_{x_1}^{x_2} F(x) \, dx$

   In this case, $F(x) = -kx = -10x$ N, $x_1 = 0.1$ m, and $x_2 = 1.9$ m. So,

   $W = \int_{0.1}^{1.9} -10x \, dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9} = -5 \left[ (1.9)^2 - (0.1)^2 \right] = -5 [3.61 - 0.01] = -5[3.6] = -18 \, \text{J}$

2. Apply the work-energy theorem:

   The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy:

   $W = \Delta KE = KE_f - KE_i$

   Where $KE_i = \frac{1}{2}mv_i^2$ is the initial kinetic energy and $KE_f = \frac{1}{2}mv_f^2$ is the final kinetic energy.

   We have $m = 1$ kg, $v_i = 10$ m/s, and $W = -18$ J. Thus,

   $-18 = \frac{1}{2}(1)v_f^2 - \frac{1}{2}(1)(10)^2$

   $-18 = \frac{1}{2}v_f^2 - 50$

   $\frac{1}{2}v_f^2 = 50 - 18 = 32$

   $v_f^2 = 64$

   $v_f = \sqrt{64} = 8 \, \text{m/s}$

Therefore, the final speed of the block as it crosses the rough region is 8 m/s.