Question

chuck wagon applies a 500 - n force to accelerate a 50 - kg box. the force of friction acting on the box is 250 - n.
f_{norm}=______n
f_{frict}=______n
f_{app}=______n
f_{grav}=______n
m =______kg
f_{net}=______n
a =______m/s/s

Explanation:

Step1: Determine \(F_{frict}\)

Given \(F_{frict} = 250\space N\) (from problem statement)

Step2: Determine \(F_{app}\)

Given \(F_{app}=500\space N\) (from problem statement)

Step3: Calculate \(F_{grav}\)

Using \(F_{grav}=mg\), where \(m = 50\space kg\) and \(g = 9.8\space m/s^{2}\)
\(F_{grav}=50\times9.8=490\space N\)

Step4: Calculate \(F_{norm}\)

Since there is no vertical acceleration (\(a_y = 0\)), \(F_{norm}=F_{grav}\) (by Newton's second law in vertical direction \(F_{net,y}=ma_y = 0\)), so \(F_{norm}=490\space N\)

Step5: Calculate \(F_{net}\)

Using \(F_{net}=F_{app}-F_{frict}\)
\(F_{net}=500 - 250=250\space N\)

Step6: Calculate \(a\)

Using \(F_{net}=ma\), so \(a=\frac{F_{net}}{m}\)
\(a=\frac{250}{50}=5\space m/s^{2}\)

Answer:

\(F_{frict}=250\space N\)

\(F_{app}=500\space N\)

\(F_{grav}=490\space N\)

\(F_{norm}=490\space N\)

\(m = 50\space kg\)

\(F_{net}=250\space N\)

\(a = 5\space m/s^{2}\)