Question

analyze this: a rightward force is applied to a 7.49 - kg object to accelerate it to the right at 19.4 m/s/s. the coefficient of friction for the object and the surface is 0.585. complete the diagram.
tap on a field to enter or edit its value.
units
force: n
mass: kg
acceln: m/s/s
μ: no unit
color key:

Explanation:

Step1: Calculate \(F_{grav}\)

Using \(F_{grav}=mg\), where \(m = 7.49\space kg\) and \(g = 9.8\space m/s^{2}\)
\(F_{grav}=7.49\times9.8 = 73.402\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\)
\(F_{norm}=73.402\space N\)

Step3: Find \(F_{frict}\)

Using \(F_{frict}=\mu F_{norm}\), with \(\mu = 0.585\)
\(F_{frict}=0.585\times73.402=42.94017\space N\)

Step4: Calculate \(F_{net}\)

Using \(F_{net}=ma\), where \(m = 7.49\space kg\) and \(a = 19.4\space m/s^{2}\)
\(F_{net}=7.49\times19.4 = 145.306\space N\)

Step5: Solve for \(F_{app}\)

Using \(F_{net}=F_{app}-F_{frict}\), so \(F_{app}=F_{net}+F_{frict}\)
\(F_{app}=145.306 + 42.94017=188.24617\space N\)

Answer:

\(F_{norm}=73.4\space N\), \(F_{grav}=73.4\space N\), \(F_{frict}=42.9\space N\), \(F_{app}=188.2\space N\), \(m = 7.49\space kg\), \(a = 19.4\space m/s^{2}\), \(F_{net}=145.3\space N\), \(\mu = 0.585\)