You drop a 2.30 kg textbook to a friend who stands on the ground at distance D =

Question

You drop a 2.30 kg textbook to a friend who stands on the ground at distance D = 10.0 m below. Your friend's outstretched hands are at distance d = 1.50 m above the ground. How much work Wg does the gravitational force do on the book as it drops to her hands? J What is the change deltaU in the gravitational potential energy of the textbook-Earth system during the drop? J If the gravitational potential energy of that system is zero at ground level, what is its potential energy U when the textbook is released? J If the gravitational potential energy of that system is zero at ground level, what is its potential energy U when the textbook reaches the hands? J How much work Wg is done on the textbook by its weight as it drops to your friend's hands if U is 100 J at the ground level. J What is the change deltaU in the gravitational potential energy of the textbook-Earth system during the drop if U is 100 J at the ground level. J Find U at the release point when U is 100 J at the ground level. J Find U at the hands when U is 100 J at the ground level. J

Explanation / Answer

a) In this case, we consider the distance between the two friendsas the book ends in the other friend's hand at d. Gravitationalenergy change is given Us = mghf -mgh0= (2.3)(9.81)(1.5-1.5) -(2.3)(9.81)(10-1.5)=-191.79 J. b) The gravitational energy change with respect of the Earth takingthe same case as part a is Ue = mghf -mgh0 = (2.3)(9.81)(1.5)-(2.3)(9.81)(10) =-191.79 J c) The height over the ground is h = 10 m, so the potential energyis mgh = (2.3)(9.81)(10) = 225.63 J d) The height over the ground is h = 1.5, thus the potential energyover the ground is mgh = (2.3)(9.81)(1.5) = 33.8445J e) The potential energy U is 100 J, if U=mgh, then h= U/(mg) =100/(2.3)(9.81) = 4.432 m over the ground.Now, the energy done by the book's weight as it reaches the handsof the friend depends on the distance between the book and thehands, this is 4.432 - 1.5 = 2.932 and the energy is U=mgh =(2.3)(9.81)(2.932) = 66.15 J. f) Work change with respect of Earth isUe = Uf - Uo = (2.3)(9.81)(1.5) J - 100J = -66.15 J g) The release point is 10 - 4.432 = 5.568 m over the 100 J level.Then U = mgh = (2.3)(9.81)(5.568)= -125.63 J. The energy is negative since this pointis over the 100 J level. h) The distance between the 100 J level and the hands is 4.432 -1.5 = 2.932 m and U = mgh = (2.3)(9.81)(2.932)= 66.15J. If you have any question don't forget to PM-me, if you don't haveany don't forget to put lifesaver.

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