Answer :
Given,
The combined mass of the sledge and the younger sister, m=35 kg
The angle made by the rope, θ=13°
The force applied, F=270 N
The coefficient of friction, μ=0.27
The acceleration due to gravity, g=9.8 m/s²
(a) The normal force is given by,
[tex]N=mg-F\sin \theta[/tex]On substituting the known values,
[tex]\begin{gathered} N=35\times9.8-270\sin (13^{\circ}) \\ =282.3\text{ N} \end{gathered}[/tex]Thus the normal force is given by 282.3 N
(b) The net force on the sledge is
[tex]\begin{gathered} ma=F\cos \theta-f \\ =F\cos \theta-N\mu \end{gathered}[/tex]Where f is the frictional force.
Thus the acceleration will be,
[tex]a=\frac{F\cos \theta-N\mu}{m}[/tex]On substituting the known values,
[tex]\begin{gathered} a=\frac{270N\times\cos 13^{\circ}-282.3\times0.27}{35} \\ =5.3m/s^2 \end{gathered}[/tex]Thus the acceleration of the sledge is 5.3 m/s²