A 2.7-kg box is released on a horizontal surface with an initial speed of 2.9 m/s.It moves on the surface with a deceleration of 0.27 m/s^2.Calculate the kinetic friction force on the box
The friction force equals
(deceleration rate)*(mass)
= 0.27*2.7 = 0.729 Newtons
The initial speed does not matter
find the equivalent capacitance of a 4.20μF capacitor and an 8.50μF capacitor when they are connected\
To calculate the kinetic friction force on the box, we first need to determine the normal force acting on the box.
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the box is on a horizontal surface, so the normal force will be equal in magnitude and opposite in direction to the weight of the box.
The weight of the box can be calculated using the formula:
Weight = mass × gravity
Given that the mass of the box is 2.7 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the box:
Weight = 2.7 kg × 9.8 m/s^2 = 26.46 N
Therefore, the normal force acting on the box is 26.46 N.
Next, we can calculate the friction force using the formula:
Friction force = coefficient of kinetic friction × normal force
The coefficient of kinetic friction depends on the surfaces in contact. Since it is not given in the problem, let's assume a typical value of 0.3 for a rough surface.
Friction force = 0.3 × 26.46 N = 7.94 N
Therefore, the kinetic friction force acting on the box is 7.94 N.