Question

analyze this: a rightward force is applied to a 6.8 - kg object to accelerate it to the right at 19.6 m/s/s. the coefficient of friction for the object and the surface is 0.591. complete the diagram.
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units
force: n
mass: kg
acceln: m/s/s
μ: no unit

Explanation:

Step1: Calculate \(F_{grav}\)

Using \(F_{grav}=mg\) where \(m = 6.8\space kg\) and \(g = 9.8\space m/s^{2}\)
\(F_{grav}=6.8\times9.8 = 66.64\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\)
\(F_{norm}=66.64\space N\)

Step3: Calculate \(F_{frict}\)

Using \(F_{frict}=\mu F_{norm}\), \(\mu = 0.591\) and \(F_{norm}=66.64\space N\)
\(F_{frict}=0.591\times66.64\approx39.38\space N\)

Step4: Calculate \(F_{net}\)

Using \(F_{net}=ma\), \(m = 6.8\space kg\) and \(a = 19.6\space m/s^{2}\)
\(F_{net}=6.8\times19.6 = 133.28\space N\)

Step5: Calculate \(F_{app}\)

Using \(F_{net}=F_{app}-F_{frict}\), so \(F_{app}=F_{net}+F_{frict}\)
\(F_{app}=133.28 + 39.38=172.66\space N\)

Answer:

• \(m = 6.8\space kg\)
• \(a = 19.6\space m/s^{2}\)
• \(\mu=0.591\)
• \(F_{grav}=66.64\space N\)
• \(F_{norm}=66.64\space N\)
• \(F_{frict}\approx39.38\space N\)
• \(F_{net}=133.28\space N\)
• \(F_{app}=172.66\space N\)