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บทที่ 5 · งานและพลังงาน

Problem 5.17

หัวข้อ: งาน (Work)

วัตถุมวล 2.5 kg เริ่มเคลื่อนที่จากหยุดนิ่งโดยแรงดังรูปด้านล่าง ในช่วง 3.0 วินาที แรงนี้ทำงานเท่าใด? F (N) 10.0 t (s) 2.00 4.00 -5.00
คำตอบ
\(W =\frac{(F_{1}t_{1} + F_{2}t_{2})^{2}}{2m}= 45 J \;\blacksquare\)
ดูวิธีทำ

กำหนดให้ \(m = 4.00 kg, F_{1} = 10.0 N, t_{1} = 2.00 s, F_{2} = -5.00 N,\)

\(t_{2} = 3.00 - 2.00 s = 1.00 s, u_{1} = 0 m/s, u_{2} = v_{1} = u_{1} + a_{1}t_{1}\)

จาก \(W = W_{1} + W_{2}\)

\(W = \vec{F}_{1}\cdot \Delta \vec{r}_{1} + \vec{F}_{2}\cdot \Delta \vec{r}_{2}\)

\(W=F_{1}(\vec{u}_{1}t_{1}+\frac{1}{2}\vec{a}_{1}t_{1}^{2})+F_{2}(\vec{u}_{2}t_{2}+\frac{1}{2}\vec{a}_{2}t_{2}^{2})\)

\(W=F_{1}(\frac{1}{2}\vec{a}_{1}t_{1}^{2})+F_{2}(\vec{u}_{2}t_{2}+\frac{1}{2}\vec{a}_{2}t_{2}^{2});u_{1}=0m/s\)

\(W=F_{1}(\frac{1}{2}(\frac{F_{1}}{m})t_{1}^{2})+F_{2}((u_{1}+a_{1}t_{1})t_{2}+\frac{1}{2}(\frac{F_{2}}{m})t_{2}^{2})\)

\(W =\frac{F_{1}^{2}t_{1}^{2}}{2m}+ F_{2}((\frac{F_{1}}{m})t_{1}t_{2}+\frac{1}{2}(\frac{F_{2}}{m})t_{2}^{2}) ; u_{1} = 0 m/s\)

\(W =\frac{F_{1}^{2}t_{1}^{2} + 2F_{1}t_{1}F_{2}t_{2} + F_{2}^{2}t_{2}^{2}}{2m}\)

\(W =\frac{(F_{1}t_{1} + F_{2}t_{2})^{2}}{2m}= 45 J \;\blacksquare\)

โจทย์อื่นในหัวข้อ งาน (Work)
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