Two people push on a box that has a mass of 10kg, the person on the right has a mass of 86kg and applies a horizontal force of 50N. The person on the left has a mass of 45kg. The coefficient of kinetic friction between the box and the floor is .2, the coefficient of static friction is .4. If the box moves to the left with a constant velocity of 10 m/s, what is the force applied by the person on the left?
To find the force applied by the person on the left, we need to consider the net force acting on the box.
First, let's calculate the force of friction acting on the box. We know that the coefficient of kinetic friction between the box and the floor is 0.2. The force of kinetic friction (Fk) can be calculated using the equation:
Fk = μk * N
where μk is the coefficient of kinetic friction and N is the normal force.
The normal force (N) is the force exerted by the floor on the box and is equal to the weight of the box (mg), where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = mg
Substituting the known values:
m = 10 kg
g ≈ 9.8 m/s^2
N = (10 kg) * (9.8 m/s^2)
N = 98 N
Now, we can calculate the force of kinetic friction:
Fk = (0.2) * (98 N)
Fk = 19.6 N
Since the box is moving with a constant velocity, the net force acting on the box must be zero. This means that the force applied by the person on the left is equal in magnitude but opposite in direction to the force of friction:
Force applied by the person on the left = - Fk
Force applied by the person on the left = - 19.6 N
Therefore, the force applied by the person on the left is approximately 19.6 Newtons in the opposite direction of motion.