Answer:
[tex]F=120.96N[/tex]
Explanation:
Given the information from the exercise we need to use Newton's Laws to solve it
If we analyze the y-axis
∑[tex]F_{y}=N-W=0[/tex]
[tex]N=W=m*g[/tex]
Since the trunk is moving in the x-axis and its velocity is changing with time, its acceleration is:
[tex]a=\frac{10m/s}{20s}=0.5m/s^2[/tex]
Knowing that we can analyze the forces acting in the horizontal direction
∑[tex]F_{x}=F-fr=m*a[/tex]
[tex]F=m*a+fr[/tex]
[tex]F=m*a+uN=m*a+u*m*g[/tex]
[tex]F=(22kg)(0.5m/s^2)+(0.51)(22kg)(9.8m/s^2)=120.96N[/tex]
So, the force required to push the trunk is 120.96 N
