A block of mass 1 kg, moving along x with speed vi = 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 Fr = -kx, with k = 10 N/m. Then the final speed of the block as it crosses the rough region is:
- 4 m/s
- 10 m/s
- 8 m/s
- 6 m/s
Answer
(3) 8 m/s
Solution
Work done by the retarding (variable) force:
W = 0.11.9 (-kx) dx = - k2 (x22 - x12 )
W = - 102 (1.92 - 0.12 ) = -5(3.61 - 0.01) = -18 J
Initial kinetic energy: Ki = 12 1 102 = 50 J
Work–energy theorem: Kf = Ki + W = 50 - 18 = 32 J
12 1 vf2 = 32
vf = 8 m/s
Answer: (3)