Question

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analyze this: a rightward force of 348 n is applied to a 26.9 - kg object to accelerate it to the right. the coefficient of friction for the object and the surface is 0.571. 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 \(m = 26.9\space kg\) and \(g = 9.8\space m/s^{2}\).
\(F_{grav}=26.9\times9.8 = 263.62\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration (\(a_y = 0\)), \(F_{norm}=F_{grav}\).
\(F_{norm}=263.62\space N\)

Step3: Calculate \(F_{frict}\)

Using \(F_{frict}=\mu F_{norm}\), with \(\mu = 0.571\) and \(F_{norm}=263.62\space N\).
\(F_{frict}=0.571\times263.62\approx150.53\space N\)

Step4: Find \(F_{net}\)

\(F_{net}=F_{app}-F_{frict}\), \(F_{app} = 348\space N\) and \(F_{frict}\approx150.53\space N\)
\(F_{net}=348 - 150.53=197.47\space N\)

Step5: Calculate acceleration \(a\)

Using \(F_{net}=ma\), so \(a=\frac{F_{net}}{m}\)
\(a=\frac{197.47}{26.9}\approx7.34\space m/s^{2}\)

Answer:

\(F_{norm}=263.62\space N\), \(F_{grav}=263.62\space N\), \(F_{frict}\approx150.53\space N\), \(F_{net}=197.47\space N\), \(a\approx7.34\space m/s^{2}\), \(\mu = 0.571\) (already given), \(m = 26.9\space kg\) (already given), \(F_{app}=348\space N\) (already given)