Answer:
The speed is [tex]v =8.17 m/s[/tex]
Explanation:
From the question we are told that
The angle of slant is [tex]\theta = 37.0^o[/tex]
The weight of the toolbox is [tex]W_t = 92.0N[/tex]
The mass of the toolbox is [tex]m = \frac{92}{9.8} = 9.286kg[/tex]
The start point is [tex]d = 4.25m[/tex] from lower edge of roof
The kinetic frictional force is [tex]F_f = 22.0N[/tex]
Generally the net work done on this tool box can be mathematically represented as
[tex]Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction[/tex]
The workdone due to weigh is = [tex]mgsin \theta * d[/tex]
The workdone due to friction is = [tex]F_f \ cos\theta * d[/tex]
Substituting this into the equation for net workdone
[tex]W_{net} = mgsin\theta * d + F_f \ cos \theta *d[/tex]
Substituting values
[tex]W_{net} = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25[/tex]
[tex]= 309.98 J[/tex]
According to work energy theorem
[tex]W_{net} = \Delta Kinetic \ Energy[/tex]
[tex]W_{net} = \frac{1}{2} m (v - u)^2[/tex]
From the question we are told that it started from rest so u = 0 m/s
[tex]W_{net} = \frac{1}{2} * m v^2[/tex]
Making v the subject
[tex]v = \sqrt{\frac{2 W_{net}}{m} }[/tex]
Substituting value
[tex]v = \sqrt{\frac{2 * 309.98}{9.286} }[/tex]
[tex]v =8.17 m/s[/tex]
