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Jbum
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1. a table is being sanded. in the process, the sandpaper is rubbed back and forth 30 times over a distance of 0.6m. it is pressed against the table with a normal force of 1.5N, and the coefficient of kinetic friction is 0.85. how much work is being done by the kinetic frictional force during this process? 2. W = Fd and f = N(uk)3. force of friction = 1.5N (0.85) = 1.275N
therefore, work = 1.275N (0.6m) = 0.765J
is this correctly done??and one more additional question as a side note: is it correct to say that no work is don't if the displacement from the original position is zero??thanks for the help.
therefore, work = 1.275N (0.6m) = 0.765J
is this correctly done??and one more additional question as a side note: is it correct to say that no work is don't if the displacement from the original position is zero??thanks for the help.
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