Answer:
[tex]P=28.085\,hp[/tex]
Explanation:
Given that:
Now, force normal to the inclined plane:
[tex]F_N=m.g.cos\theta[/tex]
[tex]F_N=80\times 9.8\times cos25^{\circ}[/tex]
[tex]F_N=710.54\,N[/tex]
Frictional force:
[tex]f=\mu.F_N[/tex]
[tex]f=0.15\times 710.54[/tex]
[tex]f=106.58\,N[/tex]
The component of weight along the inclined plane:
[tex]W_l=m.g.sin\theta[/tex]
[tex]W_l=80\times 9.8\times sin25^{\circ}[/tex]
[tex]W_l=331.33\,N[/tex]
Now the total force required along the inclination to move at the top of hill:
[tex]F=f+W_l[/tex]
[tex]F=106.58+331.33[/tex]
[tex]F=437.91\,N[/tex]
Hence the work done:
[tex]W=F.l[/tex]
[tex]W=437.91\times 220[/tex]
[tex]W=96340.80\,J[/tex]
Now power:
[tex]P=\frac{W}{t}[/tex]
[tex]P=\frac{96340.80}{138}[/tex]
[tex]P=698.12\,W[/tex]
So, power required for 30 such bodies:
[tex]P=30\times 698.12[/tex]
[tex]P=20943.65\,W[/tex]
[tex]P=\frac{20943.65}{745.7}[/tex]
[tex]P=28.085\,hp[/tex]