it takes a force of 22 N to push a 10kg box horizontally with a constant velocity over concrete. what is the coeffcient of kinetic friction between the box and the concrete
To find the coefficient of kinetic friction, we first need to calculate the normal force acting on the box. The normal force is equal in magnitude and opposite in direction to the force of gravity acting on the box.
The force of gravity on the box can be calculated using the formula: F_gravity = m * g
where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 10 kg * 9.8 m/s^2 = 98 N
Since the box is moving at a constant velocity, the frictional force opposing its motion must be equal in magnitude and opposite in direction to the applied force. Therefore, the force of friction is also 22 N.
The coefficient of kinetic friction (μ_k) can be found using the formula: F_friction = μ_k * F_normal
Since the box is on a horizontal surface, the normal force is equal to the force of gravity (F_normal = 98 N).
22 N = μ_k * 98 N
Dividing both sides of the equation by 98 N:
μ_k = 22 N / 98 N
μ_k = 0.2245
Therefore, the coefficient of kinetic friction between the box and the concrete surface is approximately 0.2245.