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?

1 year ago

$F_{net}=m.a$F_{net}=m.a => $a=\frac{F_{net}}{m}=\frac{\left(F_{app}-F_f\right)}{m}=\frac{\left(150N-50N\right)}{25kg}=4$a=F_{net}m =(F_{app}−F_{ƒ })m =(150N−50N)25kg =4$\frac{m}{s^2}$ms^{2} [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 = F_{net/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.}

1 year ago

## Answered By Majid B

$F_{net}=m.a$F

_{net}=m.a => $a=\frac{F_{net}}{m}=\frac{\left(F_{app}-F_f\right)}{m}=\frac{\left(150N-50N\right)}{25kg}=4$a=F_{net}m =(F_{app}−F_{ƒ })m =(150N−50N)25kg =4$\frac{m}{s^2}$ms^{2}[E]1 year 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 aa = F

_{net/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.}