Asked by Kayla
A 15 kg box is given an initial push so that it slides across the floor & comes to a stop. If the coefficient of friction is .30,
A) Find the Ffr Fg=mg=Fg=15(9.8)=147N
Ffr=mu(Fn)- Ffr=.3*147N=44.1N
B)Find the acceleration. Hint: what is the net force as the box slides to a stop?
I want to say 0 but then C) asks what d will be if it's initial speed is 3.0 m/s & I can't get any answer other than 0 if a & V. are 0...
A) Find the Ffr Fg=mg=Fg=15(9.8)=147N
Ffr=mu(Fn)- Ffr=.3*147N=44.1N
B)Find the acceleration. Hint: what is the net force as the box slides to a stop?
I want to say 0 but then C) asks what d will be if it's initial speed is 3.0 m/s & I can't get any answer other than 0 if a & V. are 0...
Answers
Answered by
bobpursley
the force of friction is 15*9.8*.3
acceleration= forcefriction/mass
acceleration= forcefriction/mass
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