7-6 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0&=0\\ m&=11.0\ut{kg}\\ F&=7.00\ut{N} \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$a=\frac{F}{m}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} d&=(0)t+\frac{1}{2}\(\frac{F}{m}\)t^2\\ &=\frac{Ft^2}{2m} \end{aligned} $$ $$ \begin{aligned} W(t)&=\vec F \cdot \vec d,\\ &=\frac{F^2t^2}{2m}\\ &=\frac{49}{22}t^2\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} W(1)&=\frac{49}{22}\ut{J}\\ &\approx 2.227272727272727\ut{J}\\ &\approx 2.23\ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W(2)&=\frac{98}{11}\ut{J}\\ &\approx 8.909090909090908\ut{J}\\ &\approx 8.91\ut{J}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} W(3)&=\frac{441}{22}\ut{J}\\ &\approx 20.045454545454547\ut{J}\\ &\approx 20.0\ut{J}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} P(t)&=\dxt{W}\\ &=\dt\(\frac{49}{22}t^2\)\\ &=\frac{49}{11}t\\ \end{aligned} $$ $$ \begin{aligned} P(3)&=\frac{147}{11}\ut{W}\\ &\approx 13.363636363636363 \ut{W}\\ &\approx 13.4 \ut{W}\\ \end{aligned} $$