Two forces parallel to the x axis do 17.1 J of work on a small tray...
Two forces parallel to the x axis do 17.1 J of work on a small tray while moving it 24.4 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?
The answer is -3.23N,but I cannot figure out how to reach this.
Homework Answers
total work w = 17.1 J
let the two parallele force be F1 = ? and F2 = 3.93 N
net force = F1 + F2
displacement s = 24.4m
W = (F1 + F2) s
17.1 = ( F1 + 3.93) 24.4
F1 = - 3.23 N
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