A block of mass m = 1 kg moving on a horizontal surface with speed vi=2ms−1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range. Fr=−kx for 0.1<x<2.01. Fr=0 for x<0.1m and x>2.01m. What is the final speed(ms) of the block as it crosses the path? (Given: log 20.1 = 1.30)

A block of mass m = 1 kg moving on a horizontal surface with speed vi=

1.	A block of mass m = 1 kg moving on a horizontal surface with speed vi=2ms−1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range. Fr=−kx for 0.1<x<2.01. Fr=0 for x<0.1m and x>2.01m. What is the final speed(ms) of the block as it crosses the path? (Given: log 20.1 = 1.30)
Answer» A block of mass m = 1 kg moving on a horizontal surface with speed $v_{i} = 2 m s^{- 1}$ enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force $F_{r}$ on the block in this range is inversely proportional to x over this range. $F_{r} = - \frac{k}{x}$ for $0.1 < x < 2.01$ . $F_{r} = 0$ for $x < 0.1 m$ and $x > 2.01 m$ . What is the final speed $(\frac{m}{s})$ of the block as it crosses the path? (Given: log 20.1 = 1.30)

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