For the part a) we need only the momentum of the box and we have the data to find it.
Momentum is given by,
[tex]p=mv[/tex]
where clearly, p is the momentum, m the mass of the box and v is the velocity.
Substituting,
[tex]p=(5kg)(5m/s)\\p=25kg.m/s[/tex]
For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,
[tex]p=(5kg)(5m/s)\\p=25kg.m/s[/tex]
It is the same that part a)
(a) The momentum of the box with respect to the floor is 25 kgm/s.
(b) The momentum of the box with respect to the floor after she puts the box down on the frictionless skating surface is 25 kgm/s.
The momentum of the box with respect to the floor is calculated as follows;
Pi = mv
where;
Pi = 5 x 5
Pi = 25 kgm/s
The momentum of the box with respect to the floor after she puts the box down on the frictionless skating surface is determined by applying the principle of conservation of linear momentum.
Pf = Pi
where;
Thus, the momentum of the box with respect to the floor after she puts the box down on the frictionless skating surface is 25 kgm/s.
Learn more about conservation of linear momentum here: https://brainly.com/question/7538238
