Question

analyze this: a frictional force acts upon a 30.1 kg rightward - moving box to accelerate it leftward at 6.88 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 \(F_{grav}\)

Using \(F_{grav}=mg\), where \(m = 30.1\space kg\) and \(g = 9.8\space m/s^{2}\).
\(F_{grav}=30.1\times9.8=294.98\space N\)

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 \(F_{norm}=F_{grav}\).
\(F_{norm}=294.98\space N\)

Step3: Calculate \(F_{net}\)

Using \(F_{net}=ma\), with \(m = 30.1\space kg\) and \(a=6.88\space m/s^{2}\) (leftward, so take magnitude).
\(F_{net}=30.1\times6.88 = 207.088\space N\)

Step4: Find \(F_{frict}\)

Since \(F_{net}=F_{frict}\) (horizontal direction, no other horizontal forces), \(F_{frict}=207.088\space N\)

Answer:

\(m = 30.1\space kg\)
\(a = 6.88\space m/s^{2}\)
\(F_{grav}=294.98\space N\)
\(F_{norm}=294.98\space N\)
\(F_{frict}=207.088\space N\)
\(F_{net}=207.088\space N\)