The impulse the floor exert on the box is 116 kgm/s.
The given parameters;
The final velocity of the box when it dropped to the floor is calculated as follows;
[tex]\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 1.1} \\\\v = 4.64 \ m/s[/tex]
The impulse the floor exert on the box is calculated as follows;
the impulse the floor exert on the box is equal to change in momentum of the book
[tex]J = \Delta P\\\\J = \Delta Mv\\\\J = M(v_f - v_0)\\\\J = 25(4.64 - 0)\\\\J = 116 \ kgm/s[/tex]
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