Question

analyze this: a frictional force of 115.5 n acts upon a 14.05 kg rightward - moving box to accelerate it leftward. complete the diagram.
tap on a field to enter or edit its value.
units
force: n
mass: kg
acceln: m/s/s

Explanation:

Step1: Determine the net force

The net force \(F_{net}\) is given as \(115.5\space N\) (since the frictional force is the only un - balanced force acting on the box in the horizontal direction as there is no other horizontal force mentioned and the vertical forces \(F_{norm}\) and \(F_{grav}\) cancel each other out (\(F_{norm}=F_{grav} = 137.69\space N\))).

Step2: Calculate the acceleration using Newton's second law

Newton's second law is \(F_{net}=ma\). We know \(m = 14.05\space kg\) and \(F_{net}=115.5\space N\). To find \(a\), we use the formula \(a=\frac{F_{net}}{m}\).
Substitute the values: \(a=\frac{115.5}{14.05}\approx8.22\space m/s^{2}\)

Answer:

The values in the diagram are consistent with the calculations. The frictional force \(F_{frict}\) (which is the net force \(F_{net}\)) is \(115.5\space N\), the mass \(m = 14.05\space kg\), the acceleration \(a=\frac{115.5}{14.05}\approx8.22\space m/s^{2}\), the normal force \(F_{norm}=mg\) (where \(g = 9.8\space m/s^{2}\), \(F_{norm}=14.05\times9.8 = 137.69\space N\)) and the gravitational force \(F_{grav}=mg=14.05\times9.8 = 137.69\space N\)