Complete Question
Force of 92 N is exerted horizontally on an 18 kg box of books, initially at rest , 7.6 m across a floor . The coefficient of kinetic friction between the floor and the box is 0.35. Find the net work done on the box.
Answer:
The value is [tex]W_{net} = 230 \ J[/tex]
Explanation:
From the question we are told that
The force exerted is [tex]F = 92 \ N[/tex]
The mass of the box is [tex]m = 18 \ kg[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.35[/tex]
The distance covered is [tex]d = 7.6 \ m[/tex]
Generally net workdone is mathematically represented as
[tex]W_{net} = F_{net} * d[/tex]
Here [tex]F_{net}[/tex] is the net force experienced by the box of books which is mathematically represented as
[tex]F_{net} = F - \mu_k * m * g[/tex]
=> [tex]F_{net} = 30.26 \ N[/tex]
So
[tex]W_{net} = 30.26 * 7.6[/tex]
=> [tex]W_{net} = 230 \ J[/tex]
