Question

having problems staying logged in or are you experiencing issues? please visit our troubleshooting section for solutions. analyze this: a frictional force acts upon a 12.74 kg rightward - moving box to accelerate it leftward at 5.78 m/s/s. complete the diagram. tap on a field to enter or edit its value. units force: n mass: kg acceln: m/s/s

Explanation:

Step1: Calculate gravitational force

Using formula \(F_{grav}=mg\), where \(m = 12.74\space kg\) and \(g = 9.8\space m/s^{2}\)
\(F_{grav}=12.74\times9.8 = 124.852\space N\)

Step2: Calculate normal force

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\)
\(F_{norm}=124.852\space N\)

Step3: Calculate net force

Using formula \(F_{net}=ma\), where \(m = 12.74\space kg\) and \(a=- 5.78\space m/s^{2}\) (leftward)
\(F_{net}=12.74\times(-5.78)=-73.6372\space N\)

Step4: Calculate frictional force

Since \(F_{net}=F_{frict}\) (only horizontal force is friction)
\(F_{frict}=73.6372\space N\) (magnitude, direction is left - ward as given)

Answer:

\(F_{grav}=124.852\space N\), \(F_{norm}=124.852\space N\), \(F_{frict}=73.6372\space N\), \(m = 12.74\space kg\), \(a=-5.78\space m/s^{2}\), \(F_{net}=-73.6372\space N\)