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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, the 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


Respuesta :

Kalahira
B) 14.0 N    

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".

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