Given data:
* The force applied on the box is,
[tex]F_a=800\text{ N}[/tex]* The mass of the box is m = 35 kg.
* The distance traveled by the box is d = 14 m.
* The coefficient of kinetic friction is,
[tex]\mu_k=0.4[/tex]Solution:
The normal force acting on the box is,
[tex]F_N=mg[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} F_N=35\times9.8 \\ F_N=343\text{ N} \end{gathered}[/tex]The kinetic frictional force acting on the box is,
[tex]F_k=\mu_kF_N[/tex]Substituting the known values,
[tex]\begin{gathered} F_k=0.4\times343 \\ F_k=137.2\text{ N} \end{gathered}[/tex]The net force acting on the box in terms of applied force and frictional force is,
[tex]\begin{gathered} F_{\text{net}}=F_a-F_k \\ F_{\text{net}}=800-137.2 \\ F_{\text{net}}=662.8\text{ N} \end{gathered}[/tex]Thus, the work done in accelerating the box is,
[tex]\begin{gathered} W=F_{\text{net}}\times d \\ W=662.8\times14 \\ W=9279.2\text{ J} \end{gathered}[/tex]Thus, the work done in accelerating the box is 9279.2 J or approximately 9.3 kJ.
