A young girl with mass 40.0 kg is sliding on a horizontal, frictionless surface

Question

A young girl with mass 40.0 kg is sliding on a horizontal, frictionless surface with an initial momentum that is due east and that has a magnitude of 90.0 kgm/s. Starting at t=0 , a net force with magnitude
F=(8.20 N/s)t
and direction due west is applied to the girl. (a) At what value of t does the girl have a westward momentum of magnitude 60.0 kgm/s? (b) How much work has been done on the girl by the force in the time interval from t= 0 to the time calculated in part (a)? (c) What is the magnitude of the acceleration of the girl at the time calculated in part (a)?

Explanation / Answer

NOTE : i can take the force function from your other post. ......................................................................................... force F = (8.2)t N/s change in momentum P = [90 i^ - (-60 i^)] kg.m/s                                        = 150 i^ kg.m/s magnitude of the momentum P = 150 kg.m/s according to Newton's second law of motion ,     force F = P/t        P = F.t        150 = (8.2)t[t - 0]         8.2 t2 = 150     time t = 4.27 s ............................................................. the relation between momentum and kinetic energy is            K.E = P2/2m from work energy theorem , workdone W = change in K.E                       = P22/2m - P12/2m                        = [(60)2/2(40)] - [(90)2/2(40)]                        = -56.25 J ............................................................ acceleration a = F/m                      = (8.2)t / m                      = 8.2*4.27 / 40                      = 0.875 m/s2

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