A crate of bananas with a mass of 25.0 kg is dragged across a level floor by an applied force of +150 N [E] against a frictional force of 50.0 N [W]. What is the observed acceleration of the crate?
4 years ago
$F_{net}=m.a$Fnet=m.a => $a=\frac{F_{net}}{m}=\frac{\left(F_{app}-F_f\right)}{m}=\frac{\left(150N-50N\right)}{25kg}=4$a=Fnetm =(Fapp−Fƒ )m =(150N−50N)25kg =4$\frac{m}{s^2}$ms2 [E]
This is a example of Newton's Second Law of Motion
F=ma
Our net force is 150 N - 50 N = 100 N [E]
since the Force is acting East and the friction is in the opposite direction
F net = ma isolating a
a = Fnet/m substituting our values
a = 100 N/25 kg = (100 kg m/s2)/25 kg = 4 m/s2 [E]
additional things you could calculate related to this problem would be the co-efficient of friction.
4 years ago
Answered By Majid B
$F_{net}=m.a$Fnet=m.a => $a=\frac{F_{net}}{m}=\frac{\left(F_{app}-F_f\right)}{m}=\frac{\left(150N-50N\right)}{25kg}=4$a=Fnetm =(Fapp−Fƒ )m =(150N−50N)25kg =4$\frac{m}{s^2}$ms2 [E]
4 years ago
Answered By Albert S
This is a example of Newton's Second Law of Motion
F=ma
Our net force is 150 N - 50 N = 100 N [E]
since the Force is acting East and the friction is in the opposite direction
F net = ma isolating a
a = Fnet/m substituting our values
a = 100 N/25 kg = (100 kg m/s2)/25 kg = 4 m/s2 [E]
additional things you could calculate related to this problem would be the co-efficient of friction.