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Example 24 A block of mass m = 1 kgmoving on a horizontal surface with speedVi = 2 ms- enters a rough patch ranging fromx = 0·10 m to x = 2:01 m. The retarding force Fron the block in this range is inversely proportionalto x over this rangeXF. =- for 0:1 < x < 2:01 m= 0 for x < 0.1 m and x > 2:01 mwhere k = 0.5 J. What is the final K.E. andspeed Vf of the block as it crosses the patch ? |
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Answer» Hey mate..,.,,..... here is ur answer........... F = ma..mv dv/dx = -k/x = -0.5/x vdv = -0.5 dx/x v2 / 2 - (2) 2/ 2 = 0.5 ( -1n . 2.01/0.1) = -1.5 v2 = 2 ( 2- 1.5) = 1 v = 1 m/s . hope it helps♥♥♥ |
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