A 50.0-kg box is sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the box is sliding to the right at 1.5 m/s and that it stops in 2.5 s with uniform acceleration. What magnitude force does friction exert on this box?
To find the magnitude of the force of friction acting on the box, we can use Newton's second law of motion. According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the force of friction.
First, let's determine the acceleration of the box. We know that the box stops in 2.5 seconds with uniform acceleration. To find the acceleration, we need to divide the change in velocity by the time taken:
Change in velocity (Δv) = final velocity (vf) - initial velocity (vi) = 0 - 1.5 m/s = -1.5 m/s
Time taken (t) = 2.5 s
Acceleration (a) = Δv / t = -1.5 m/s / 2.5 s = -0.6 m/s² (since the acceleration is negative, indicating deceleration)
Now that we have the acceleration of the box, we can find the force of friction using Newton's second law:
Net force (Fnet) = mass (m) * acceleration (a)
Given:
Mass of the box (m) = 50.0 kg
Acceleration (a) = -0.6 m/s²
Plugging in the values:
Fnet = 50.0 kg * -0.6 m/s²
Fnet = -30.0 N
The negative sign indicates that the force of friction is acting opposite to the direction of motion. Since we are interested in the magnitude of the force, we can ignore the negative sign:
Magnitude of force of friction = |Fnet| = |-30.0 N| = 30.0 N
Therefore, the magnitude of force that friction exerts on this box is 30.0 Newtons.