3 $\vect{F}_1 = (-2.00\vect{i} + 2.00\vect{j})\U{N}$,
4 $\vect{F}_2 = ( 5.00\vect{i} - 3.00\vect{j})\U{N}$, and
5 $\vect{F}_3 = -45.0\vect{i}\U{N}$,
6 act on an object to give it an acceleration of magnitude
8 \Part{a} What is the direction of the acceleration?
9 \Part{b} What is the mass of the object?
10 \Part{c} If the object is initially at rest, what is its speed $v$
11 after $t = 10.0\U{s}$?
12 \Part{d} What are the velocity components of the object after
14 \end{problem*} % problem 4.8
18 Summing the forces we have
20 \sum F_x &= F_1x + F_2x + F_3x = (-2.00 + 5.00 - 45.0)\U{N} = -42.0\U{N} \\
21 \sum F_y &= F_1y + F_2y + F_3y = (+2.00 - 3.00 + 0)\U{N} = -1.00\U{N}
23 We know from Newtons second law that
25 \sum \vect{F} = m \vect{a}
27 So the acceleration $\vect{a}$ will be in the same direction as the
29 The direction $\theta$ of the force is given by
31 \theta = \arctan \left( \frac{F_y}{F_x} \right)
32 = \arctan \left( \frac{-1}{-42} \right)
33 = (1.36 + 180)^o = \ans{181.36^o}
35 Measured counter-clockwise from the $\vect{x}$ axis
36 (where we have added $180^o$ because $F_x < 0$ so we have a backside
40 From Newton's second law
42 \sum \vect{F} &= m \vect{a} \\
43 \left|\sum \vect{F}\right| &= m \left|\vect{a}\right| \\
44 m &= \frac{\left|\sum \vect{F}\right|}{a}
45 = \frac{ \sqrt{(-41.0)^2 + (-1.00)^2}\U{N}}{3.75\U{m/s}^2}
50 This section is constant acceleration review.
53 = 3.75\U{m/s}^2 \cdot 10.0\U{s} = \ans{37.5\U{m/s}}
57 Using our velocity $v = |\vect{v}|$ from \Part{c} and our angle
58 $\theta$ from \Part{a} (we know that $\vect{v}$ is in the same
59 direction as $\vect{a}$ and $\vect{F}$) we have
61 v_x &= v \cos \theta = 37.5\U{m/s} \cdot \cos 181.36^o
63 v_y &= v \sin \theta = 37.5\U{m/s} \cdot \sin 181.36^o
65 \vect{v} &= \ans{(-37.49\vect{i} - 0.893\vect{j})\U{m/s}}