Answer:
Explanation:
Considering Work done by friction is asked
Given
mass of block [tex]m=15.4\ kg[/tex]
force [tex]F=182\ N[/tex]
inclination [tex]\theta =24^{\circ}[/tex]
block is displaced by [tex]s=57.6\ m[/tex]
coefficient of kinetic friction [tex]\mu _k=0.135[/tex]
Friction force [tex]F_r=\mu _kN[/tex]
Normal reaction [tex]N=mg-F\sin \theta =15.4\times 9.8-182\sin (24)[/tex]
[tex]N=76.9\ N[/tex]
Friction Force [tex]F_r=0.135\times 76.9=10.38\ N[/tex]
Work done by friction force
[tex]W=f_r\cdot s[/tex]
[tex]W=10.38\times 57.6=597.92\ N[/tex]