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What will be the area of contact of a block weighing 20N, if it can apply a pressure of 2000Pa? |
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Answer» Answer: Hello my FRIEND Explanation: a. F fric
=μ s
W=0.4∗20=8N b. SINCE F applied
fric F fric =5N c. MINIMUM force required to start the MOTION F=F staticfric =8N d. Minimum force required to start the motion F=F kineticfric =μ k W=0.2∗20=4N e. F>F staticfric motion started. F>F kineticfric hence motion continues. F fric =4N |
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