Answer:
The student must be pushed 8.0m.
Explanation:
Since there is an external force acting on the student, there is no energy conservation. However, if we consider the work done by the constant force, we can say that:
[tex]E_0=E_f-W[/tex]
Where E_0 is the initial energy of the student, E_f is her final energy and W is the work done by the constant force. Because the motion is in the horizontal axis, we don't have to consider the gravitational potential energy. And since the student starts from rest, her initial kinetic energy is zero. So, we have:
[tex]0=K_f-W\\\\W=K_f\\\\Fd=K_f\\\\\implies d=\frac{K_f}{F}\\\\d=\frac{355J}{44N}=8.0m[/tex]
In words, the student must be pushed 8.0m.
