An object of mass 0.550 kg is lifted from the floor to aheight of 3.50m at a constant speed. a) How much work is done by the lifting force (includeunits)? b) How much work is done by the Earth on the object? c) What is the net work done on the object? d) What is the change in kinetic energy of the object? e) Are your results consistent with the work-energy principle?Explain. An object of mass 0.550 kg is lifted from the floor to aheight of 3.50m at a constant speed. a) How much work is done by the lifting force (includeunits)? b) How much work is done by the Earth on the object? c) What is the net work done on the object? d) What is the change in kinetic energy of the object? e) Are your results consistent with the work-energy principle?Explain.
Explanation / Answer
given the mass of the object is m = 0.550kg the height it raised is h = 3.50 m a) the work done by the lifting force is W = potentialdifference = mgh = 18.865 J b) the work done by the earth on the object is W' = -mgh =-18.865 J c) the nwet work on the object is = W+W' = 0 J d) as the speed of the object is constant the change in thekinetic energy is zero . e) according to the work energy theorem work done = change in the kinetic energy 0=0