An 85-N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and floor.
2007-03-22 23:14:21 · 2 answers · asked by biscuits 2 in Science & Mathematics ➔ Physics
Ok, I have this idea :
the desacceleration is : 0.9 m/s^2
Normal component of the force = 25, then, the normal force will be equal to :
Normal force = 85 + 25 = 110 N
then, let's apply Newton's second law :
F - f = m*a
F = horizontal force = 20 N
f = force of friction = Normal*u
20 - 110u = 85/9.8*0.9
12.2 = 110u
u = 0.11 ...... Coefficient of kinetic friction
2007-03-23 05:21:54 · answer #1 · answered by anakin_louix 6 · 1⤊ 0⤋
First we'll figure out the Normal. That's going to be the "weight" of the box on the floor, including the mass and the push force. Thus, is 110N.
Next we'll figure out the force of friction: That's going to be the total acceleration (=ma, Newton's second law) minus the push force. We should get a negative result. Let's see: total acceleration is -.90*85=-76.5 (negative since it's slowing down), and push force is 25N. Thus, the force of friction is -101.5 - and it's negative.
Since the magnitude of the force of friction is equal to the coefficient of friction times the Normal, then the coefficient (mu) is the magnitude of the force of friction (so take out the negative) divided by the Normal=101.5/110=0.9227
2007-03-22 23:23:45 · answer #2 · answered by mck 2 · 0⤊ 1⤋