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applesjgtl
A 65.0 kg object is dragged with a force of 950 N at a 30º angle above a horizontal sidewalk for 1.62 km. If the coefficient of kinetic friction between the object and the sidewalk is 0.45, what is the net work performed by the person dragging the object?
Irrespective of other forces acting on the object, the work done by a particular force on an object is given by W = F.S Let us now resolve the force along the horizontal and vertical directions. Force acting in the horizontal direction = \[Fcos30 = 950*\sqrt{3} / 2\] N Force acting along the vertical direction = \[Fcos60 = 950/2 = 475N\] The vertical component of the force applied is not sufficient to lift it off the sidewalk (since it is less than its weight) . Therefore, displacement in the vertical direction = 0. Therefore, work done by the applied force = (vertical component * vertical displacement) + (horizontal component * horizontal displacement) =\[0 + 950*\sqrt{3}/2 * 1.62 * 1000\] The answer would be in joules since we have considered S.I. units throughout.
The friction applies a force too though, should that not be considered?
No. That need not be considered while calculating work done by the PERSON. If you want to calculate NET WORK done on the object, you need to consider work done by friction also
i mean... total work done on the object = net work done by person + net work done by friction