A man attempts to move a stationary object that weighs 152 N. The coefficient of static friction is .35 and the kinetic friction is .27
1. How much force must be applied to get the cabinet moving?
2. If the same force required to start the motion is applied to the cabinet after it starts to move, what is the cabinet's acceleration?
please help me!
2007-11-20 06:43:42 · 2 answers · asked by hotmvper2k3 1 in Science & Mathematics ➔ Physics
The force of static friction is P = uN = (.35)(152 N)
Solve for P; the man will have to apply that much force to get the object moving.
Once it's moving, if the man continues to apply the same force Newton's Second Law is
F = ma = P - uN where this time u is the coefficient of kinetic friction. You already solved for P; N is still 152 N, and u is .27. The mass of the cabinet is 152 N / 9.8 m/s^2 (that is, weight = mg and you already have mg). Now you can solve for a, the acceleration of the object.
2007-11-20 06:59:12 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
In order to start the cabinet moving force=(coefficient of static friction)*(normal force) ie. F=(.35)*(152N)=53.2N. After it has started moving the acceleration can be found be subtracting the starting force from the force required to keep it moving ie. F=(.27)(152N)=41.04N. That means that a force is applied which is 12.16N. Then use F=ma to find the acceleration. (Note that m=152N/(9.8m/s^2)=15.51kg.) Solve for a.
2007-11-20 15:01:00 · answer #2 · answered by jedd c 3 · 0⤊ 0⤋