Question

having problems staying logged in or are you experiencing issues? please visit our troubleshooting section for solutions analyze this: a rightward force of 60.2 n is applied to a 7.78 - kg object to accelerate it to the right at 4.71 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 μ: no unit color key: not checked correct incorrect check answers student name: wbfsmarto82@student.wbsd.org (replay - no tracking) level: wizard problem 3 of 3 health: 100% tap for question - specific help help me! view directions to main menu

Explanation:

Step1: Calculate gravitational force

Using \(F_{grav}=mg\), where \(m = 7.78\space kg\) and \(g = 9.8\space m/s^{2}\).
\(F_{grav}=7.78\times9.8 = 76.244\space N\)

Step2: Determine normal force

Since there is no vertical acceleration (\(a_y = 0\)), \(F_{norm}=F_{grav}\).
\(F_{norm}=76.244\space N\)

Step3: Find net force

Using \(F_{net}=ma\), with \(m = 7.78\space kg\) and \(a=4.71\space m/s^{2}\).
\(F_{net}=7.78\times4.71 = 36.6438\space N\)

Step4: Calculate frictional force

Using \(F_{net}=F_{app}-F_{frict}\), so \(F_{frict}=F_{app}-F_{net}\). Given \(F_{app}=60.2\space N\) and \(F_{net}=36.6438\space N\).
\(F_{frict}=60.2 - 36.6438=23.5562\space N\)

Step5: Compute coefficient of friction

Using \(F_{frict}=\mu F_{norm}\), so \(\mu=\frac{F_{frict}}{F_{norm}}\).
\(\mu=\frac{23.5562}{76.244}\approx0.309\)

Answer:

\(F_{grav} = 76.244\space N\), \(F_{norm}=76.244\space N\), \(F_{net}=36.6438\space N\), \(F_{frict}=23.5562\space N\), \(\mu\approx0.309\), \(m = 7.78\space kg\), \(a = 4.71\space m/s^{2}\), \(F_{app}=60.2\space N\)