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=- krN, with k=10 N/m. Th…

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=- krN, with k=10 N/m. Then the final speed of the block as it crosses rough region is.

Accepted Answer

Correct answer: C. (c) To find the final speed of a block moving through a rough region, consider the following steps: Given Parameters: Initial speed of the block, v_i=10 m/s. Mass of the block, m=1 kg. Force of friction acting on the block, F_r=-k x. k=10 N/m . Forces and Acceleration: The acceleration a of the block is given by a= Fm=- kx. So, we have: a=-10x. Using the Relationship: We use the relationship between velocity, acceleration, and position: v dv dx=a. Substituting a gives: v dv dx=-10x. Integrating the Equation: Integrate both sides with respect to their variables: 0.05em vdv=-10 0.05em xdx Solving the Integrals: The left side becomes v^22- 1002. The right side evaluates to -10 (1.9^2-0.1^2)2. Therefore: v^2-1002=-10 ( 1.9^2-0.1^22 ) Final Calculation: Solve the above equation for v : v^2-1002=-10 (3.61-0.01 )2 Simplifying and solving gives: v^2-100=-10 1.8 Solving further: v^2=100-18v^2=82v= 82 ≈8 m/s Therefore, the final speed of the block as it exits the rough region is approximately 8 m/s.

$A block of mass 1 kg, moving along x with speed v_{i} = 10 m / s enters a rough region ranging from x = 0.1 m to x = 1.9 m . The retarding force acting on the block in this range is F_{r} = - krN, with k = 10 N / m . Then the final speed of the block as it crosses rough region is.$

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