Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
he speed of the box was 2.25 m/s, and just after leaving it, the speed of the box was 1.20 m/s. In Situation 6.1, the magnitude of the average force that friction on the rough patch exerts on the box is closest to: A) 19.5 N
B) 14.0 N
C) 13.7 N
D) 5.55 N
E) It is impossible to know since we are not given the coefficient of kinetic friction
The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is:
E = 0.5 M V^2
where
E = Energy
M = Mass
V = velocity
Substituting the known values, let's calculate the before and after energy.
Before:
E = 0.5 M V^2
E = 0.5 13.5kg (2.25 m/s)^2
E = 6.75 kg 5.0625 m^2/s^2
E = 34.17188 kg*m^2/s^2 = 34.17188 joules
After:
E = 0.5 M V^2
E = 0.5 13.5kg (1.2 m/s)^2
E = 6.75 kg 1.44 m^2/s^2
E = 9.72 kg*m^2/s^2 = 9.72 Joules
So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N
Rounding to 1 decimal place gives 14.0 N which matches option "B".
