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A block of mass 10 kg slides down a roughslope which is inclined at 45° to the horizontal.The coefficient of sliding friction is 0.30. Whenthe block has to slide 5 m, the work done onthe block by the force of friction is nearly |
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Answer» Explanation: 4 votes F = mue × mg So, F= 0.2×5×10 F=10 Now, W= mg cos (theta) W= 10× 10 × √2\2 So , W = 50 ... More |
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