- #1
anyone1979
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Is this right?
A block of mass 1 kg is acellerated from an initial speed of 2 m/s to a final speed of 6 m/s as it is dragged a distance of 10 m over a level floor. If the coefficient of kinetic friction between the block and the floor is 0.2, how much work must the be done by the applied force?
Fk = (.2)(9.8) = 1.96N
WFk = (1.96)(10)(-1) = -19.6J
W = (1/2)(1)(6^2) - (1/2)(1)(2^2) = 16J
Wtot = 19.6 + 16 = 35.6J
A block of mass 1 kg is acellerated from an initial speed of 2 m/s to a final speed of 6 m/s as it is dragged a distance of 10 m over a level floor. If the coefficient of kinetic friction between the block and the floor is 0.2, how much work must the be done by the applied force?
Fk = (.2)(9.8) = 1.96N
WFk = (1.96)(10)(-1) = -19.6J
W = (1/2)(1)(6^2) - (1/2)(1)(2^2) = 16J
Wtot = 19.6 + 16 = 35.6J
