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: Calculate \(F_{grav}\)

Using the formula \(F_{grav}=mg\), where \(m = 14.05\space kg\) and \(g = 9.8\space m/s^{2}\).
\(F_{grav}=14.05\times9.8=137.69\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration (\(a_y = 0\)), by Newton's second law \(F_{net,y}=F_{norm}-F_{grav}=ma_y = 0\). So \(F_{norm}=F_{grav}\)
\(F_{norm}=137.69\space N\)

Step3: Calculate \(a\)

Using Newton's second law \(F_{net,x}=F_{frict}=ma_x\). Given \(F_{frict}=115.5\space N\) and \(m = 14.05\space kg\)
\(a=\frac{F_{frict}}{m}=\frac{115.5}{14.05}\approx8.22\space m/s^{2}\) (leftward, so \(a=- 8.22\space m/s^{2}\) if we take right as positive)

Step4: Determine \(F_{net}\)

Since the only horizontal force is \(F_{frict}\) (assuming no other horizontal forces), \(F_{net}=F_{frict}\)
\(F_{net}=115.5\space N\) (leftward)

Answer:

\(F_{norm}=137.69\space N\), \(F_{grav}=137.69\space N\), \(m = 14.05\space kg\), \(a\approx - 8.22\space m/s^{2}\), \(F_{net}=115.5\space N\)