In 1990 walter arfeuille of belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) how much work was done on the object by arfeuille in this lift, assuming the object was lifted at constant velocity?

For the law of conservation of energy, the work done by the teeth must be equal to the variation of mechanical energy of the object. However, there is no variation of the kinetic energy K of the object, because the velocity is constant, and so K is constant as well. So the work done is simply equal to the gravitational potential energy acquired by the object, which is equal to:

where m=281.5 kg is the mass of the object, is the gravitational acceleration and is the height at which the object has been lifted. Using these data, we find:

And as we said above, this coincides with the work done by the teeth.

Idon't know it i just need my points lol

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n