Can the total work done on an object during a displacement be negative? explain. if the total work is negative, can its magnitude be larger than the initial kinetic energy of the object? explain.

yes,work can be negative. -ve work is said to be done when the system is doing work,and +ve work when ur force is reaping results. when u apply force and get the work done in ur direction(in which u apply force) then work done is +ve. but if it does not move and stays as it was,then 0 work is done but,if there is an opposing force,and the object instead of moving in ur direction,moves in the other direction,then the work said to be done by U is in -ve but the work done by the opposing force is said to be positive work = force * the distance moved s'pose u did do some work, then the object is gonna move some distance, s'pose it is 3metres. that means u were assuming the initial position to be 0 meters. if the object was moved an oppposite distance ie. now the distance will be on the other side of the 0mark, hence -ve distance has been mived BY U., but for that opposing body it will be positive work. but in common language,,we can say only one set of worl to be done..."we cannot say work was +ve as well as -ve" hence we took UR side as human side and always said -ve or +ve work in UR respect. nowdid u understand how work came to be called -ve? see, i understood the meaning of what u r trying to say in the second question,wich i dont think my previous answeres have! anywaysi'll explain YES,its magnintude has to be larger than the initial kinetic energy. BUT ONLY IF the body is moving in the opposite direction INITIALLY. u see, to move a body in opposite direction,u'll first have to bring the energy that hlps it to move in the opposite directiion to 0. but that does not mean that the kinetic energy becomes zero. in THIS CASE,yes,there will be a fraction of a second when it will become 0, nut then again it will gain itself --- of course because its gainig speed in the opposite direction due to the opposing force

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