ANSWER and EXPLANATION
a) The free-body diagram is given below:
where N = normal force
Ff = force of friction
W = weight of the block
b) To find the net force on the body, we first have to find the components of force in the horizontal and vertical directions.
For the horizontal components of forces acting on the block, we have:
[tex]F_x=F\cos \theta-F_f[/tex]where F = force of pull
θ = angle at which the block is being pulled
and Ff is given as:
[tex]F_f=\mu\cdot m\cdot g[/tex]where m = mass; g = acceleration due to gravity
Therefore, substituting the given values into the formula:
[tex]\begin{gathered} F_x=250\cos 15-\mu m\cdot g \\ F_x=250\cos 15-(0.25\cdot90\cdot9.8) \\ F_x=241.48-220.5 \\ F_x=20.98N \end{gathered}[/tex]For the vertical components of forces acting on the block, we have:
[tex]F_y=250\sin 15+N-W[/tex]where N is given as:
[tex]N=m\cdot g[/tex]and W is given as:
[tex]m\cdot g[/tex]As we can see, the normal force and weight have the same magnitude, so they cancel out.
Hence, the vertical component of the force is:
[tex]\begin{gathered} F_y=250\sin 15 \\ F_y=64.70N \end{gathered}[/tex]The net force is the square root of the sum of the squares of the individual components. Therefore, we have that:
[tex]\begin{gathered} F=\sqrt[]{F^2_x+F^2_y} \\ F=\sqrt[]{20.98^2+64.70^2} \\ F=\sqrt[]{440.16+4186.09}=\sqrt[]{4626.25} \\ F=68.02N \end{gathered}[/tex]That is the net force acting on the block.
To find the acceleration of the block, we have to apply the formula for force, which is:
[tex]F=ma[/tex]where F = net force; m =mass of block; a = acceleration
Therefore, the acceleration of the block is:
[tex]\begin{gathered} 68.02=90\cdot a \\ \Rightarrow a=\frac{68.02}{90} \\ a=0.76m/s^2 \end{gathered}[/tex]
