Two forces parallel to the x axis do 17.2 J of work on a small tray while moving it 19.0 m in the x direction across a gym floor. One of the forces has a value of +3.93 N in the x direction. What is the other force?

Answer 1

net force*distance=work
net force= work/distance=17.2J/19m= you do it.

then the sum of forces equals net force
F1+3.93N=net force
solve for F1