Answer to Question #266713 in Classical Mechanics for usfuhF>SFJSj

Question

An ice glider is traveling 35 degrees W of S at 6 m/s when the wind exerts a force of 860 N 18 degrees N of W for three seconds. The mass of the glider is 215 kg. Complete the chart below and then answer the questions that follow.

t(s): 0, 1, 2, 3

East v (m/s): --3.44, t = 1, t = 2, t = 3

North v (m/s): --4.92, t = 1, t = 2, t = 3

What is the glider's initial kinetic energy?

What is the glider's kinetic energy at t = 2?

How much energy does the wind put into the glider in the 3 s interval?

What is the glider's average speed for the three second interval?

What is the glider's average velocity for the three second interval?

Answer 1

$v_0 = 6\frac{m}{s}$

$\text{We introduce a coordinate system with the basis: }\vec E,\vec N$

$v_{0E}= -\sin35\degree *v_0=-3.44$

$v_{0N}= -\cos35\degree *v_0=-4.92$

$F= ma$

$a = \frac{F}{m}=\frac{860}{215}= 4$

$a_{E}= -\sin18\degree *a=-1.24$

$a_{N}= -\cos18\degree *a=-3.80$

$v= v_0+at$

$v_E= v_{0Е}+a_Еt=-3.44-1.24t$

$v_N= v_{0N}+a_Nt=-4.92-3.80t$

$\begin{matrix} t(s) & 0 &1&2&3\\ v_E& -3.44&-4.68&-5.92&-7.16\\ v_N& -4.92&-8.72&-12.52&-16.32\\ \end{matrix}$

$EK_0 =\frac{mv_0^2}{2}= \frac{215*6^2}{2}= 3870J$

$v_2 = \sqrt{v_{2E}^2+v_{2N}^2}= \sqrt{5.92^2+12.52^2}=13.85\frac{m}{s}$

$EK_2 =\frac{mv_2^2}{2}= \frac{215*13.85^2}{2}= 20621J$

$v_3 = \sqrt{v_{3E}^2+v_{3N}^2}= \sqrt{7.16^2+16.32^2}=17.82\frac{m}{s}$

$EK_3 =\frac{mv_3^2}{2}= \frac{215*17.82^2}{2}= 34123J$

$\Delta EK = EK_3-EK_0= 30253J$

$v_{avg}= \frac{v_0+v_3}{3}= 7.94\frac{m}{s}$

$\text{Answer:}$

$EK_0 = 3870J$

$EK_2 =20621J$

$\Delta EK = 30253J$

$v_{avg}= 7.94\frac{m}{s}$