Answer:
[tex]v=2.48m/s[/tex]
Explanation:
From work-energy theorem we got that:
[tex]W=K_{2}-K_{1}=\frac{1}{2}m(v_{2}^2-v_{1}^2)[/tex] (1)
If we see the free body diagram attached we can calculate the work done by the group of students by knowing its definition given by:
[tex]W=F*s[/tex]
Where F is Force and s is distance
[tex]W=(Fcos(32)-wsin(32))(s)[/tex]
These are the forces at the x-axis that are causing the work done by the students
[tex]W=((599N)cos(32)-(86kg)(9.8m/s)sin(32))(2.3m)[/tex]
[tex]W=141.13J[/tex]
Now, we can calculate the velocity at the top of the ramp using (1)
[tex]W=\frac{1}{2}m(v_{2}^2-v_{1}^2)[/tex]
[tex]v_{2}=\sqrt{\frac{2W}{m}+v_{1}^2}=\sqrt{\frac{2(141.13J)}{86kg}+(1.70m/s)^2}=2.48m/s[/tex]
