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 100 kg•m/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 kg•m/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)? Please show step by step, thank you 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 100 kg•m/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 kg•m/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)? Please show step by step, thank you a) At what value of t does the girl have a westward momentum of magnitude 60.0 kg•m/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)? Please show step by step, thank you

Explanation / Answer

here,

mass of the young girl , m = 40 kg

initial momentum , Pi = 100 kg.m/s

F = (8.2 N/s ) * t

a)

the final momentum , Pf = - 60 kg.m/s

integration(F) = Pf - Pi

8.2 * t^2 /2 = 160

t = 6.25 s

the time taken by the girl is 6.25 s

b)

initial velocity , vi = P1/m

vi = 2.5 m/s

final velocity , vf = Pf/m

vf = - 1.5 m/s

using work energy theorm

work done , w = 0.5 * m * ( vf^2 - vi^2)

w = 0.5 * 40 * ( - 4)

w = - 80 J

the work done on the girl is - 80 J

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