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    				| 1. | A 30kg box has to move up an iclined slope of 309 to the horizontal at a uniform velocity d5ms1. If the frictional force retarding the motion is 150N, the horizontal force required to moveupis (g-10 ms2) | 
| Answer» I’ll assume the applied force is parallel to the slope. It has to overcome friction (150N) and the component of weight down the slope (= 300Sin30 = 150N again) So, 300N If it is applied at some other angle, we’d need to know it. And anything not parallel to the slope would affect the Normal Reaction and hence the Friction | |