Problem 5.52 Challenge

พิจารณา F.B.D. ของวัตถุมวล \(10.0 kg\)

a

\(F = 150 N\)

\(\theta\)

\(f = \mu N\)

Mg

N

จาก \(ΣF_{y}= ma_{y}\)

\(N - mg - Fsin\theta = m(0)\)

\(N = mg + Fsin\theta\)

\(N = (10.0 kg)(10 m/s^{2}) + 150\sin \theta\)

\(N = 100 + 150\sin \theta\)

จาก\(\vec{s}=\vec{u}t+\frac{1}{2}\vec{a}t^{2}\)

\(a =\frac{2(s - ut)}{t^{2}}=\frac{2(5.00 - (0)(2.00))}{(2.00)^{2}}= 2.50 m/s^{2}\)

จาก \(ΣF_{x}= ma_{x}\)

\(Fcos\theta - \mu N = ma\)

\(150\cos \theta - (0.500)(100 + 150\sin \theta ) = (10.0)(2.50)\)

\(150\cos \theta - 75.0\sin \theta - 50.0 = 25.0\)

\(6\cos \theta - 3\sin \theta - 2 = 1\)

\(2\cos \theta - 1 = \sin \theta\)

\(2\cos \theta - 1 =\sqrt{1 - \cos ^{2}\theta}\)

\(4\cos ^{2}\theta - 4\cos \theta + 1 = 1 - \cos ^{2}\theta\)

\(5\cos ^{2}\theta - 4\cos \theta = 0\)

\(\cos \theta (5\cos \theta - 4) = 0\)

\(\cos \theta =0,\frac{4}{5}\to \therefore \cos \theta =\frac{4}{5}\)

จาก \(\wp _{av}=\frac{\Delta W}{\Delta t}\)

\(\wp _{av}=\frac{\vec{F}\cdot \Delta \vec{r}}{\Delta t}=\frac{Fscos\theta}{t}=\frac{(150)(5.00)(\frac{4}{5})}{2.00}= 300 W \;\blacksquare\)