91Ó°ÊÓ

To find the net force acting on an object, we need to add up all the forces acting on it. In this case, there are three force vectors, \(\overrightarrow{\mathbf{F}}_1\), \(\overrightarrow{\mathbf{F}}_2\), and \(\overrightarrow{\mathbf{F}}_3\). We will add these vectors component-wise (i.e., add the x-components together and the y-components together), which will give us the net force vector: \[ \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{1} + \overrightarrow{\mathbf{F}}_{2} + \overrightarrow{\mathbf{F}}_{3} \]

Now we'll evaluate the net force vector components. \[ \overrightarrow{\mathbf{F}}_{net} = (-3.00 \hat{\mathbf{i}} + 2.00 \hat{\mathbf{j}}) + (6.00 \hat{\mathbf{i}} - 4.00 \hat{\mathbf{j}}) + (2.00 \hat{\mathbf{i}} + 5.00 \hat{\mathbf{j}}) \] This simplifies to: \[ \overrightarrow{\mathbf{F}}_{net} = 5.00\hat{\mathbf{i}} + 3.00\hat{\mathbf{j}}\ \mathrm{N} \]

Since we know that the magnitude of the acceleration is 4.23 m/s², we need to find the unit vector \(\hat{\mathbf{a}}\) of the acceleration to find the acceleration vector \(\overrightarrow{\mathbf{a}}\). According to Newton's second law, we know that: \[ \overrightarrow{\mathbf{F}}_{net} = m\overrightarrow{\mathbf{a}} \] Hence, we can find the unit vector \(\hat{\mathbf{a}}\) by dividing the net force vector by the magnitude of the acceleration, namely : \[ \hat{\mathbf{a}} = \frac{\overrightarrow{\mathbf{F}}_{net}}{4.23 \mathrm{m/s}^2} \] Calculate the components of \(\hat{\mathbf{a}}\) using the components of the net force vector: \[ \hat{\mathbf{a}} = \frac{5.00\hat{\mathbf{i}} + 3.00\hat{\mathbf{j}} \ \mathrm{N}}{4.23 \mathrm{m/s}^2} = 1.18\hat{\mathbf{i}} + 0.71\hat{\mathbf{j}} \]Now, we can calculate the acceleration vector: \[ \overrightarrow{\mathbf{a}} = (4.23 \mathrm{m/s}^2)(1.18\hat{\mathbf{i}} + 0.71\hat{\mathbf{j}}) = 5.00\hat{\mathbf{i}} + 3.00\hat{\mathbf{j}} \ \mathrm{m/s^2} \]

Using the net force vector and the acceleration vector, we can calculate the mass of the object with Newton's second law formulation: \[ m = \frac{|\overrightarrow{\mathbf{F}}_{net}|}{|\overrightarrow{\mathbf{a}}|} \] We have the magnitudes of the net force and acceleration vectors as the same (calculated in step 2 and step 3), so: \[ m = \frac{(\sqrt{5.00^2 + 3.00^2})\ \mathrm{N}}{(\sqrt{5.00^2 + 3.00^2})\ \mathrm{m/s^2}} \] Simplifying this expression, we can find the mass: \[ m = \frac{(\sqrt{25+9})\ \mathrm{N}}{(\sqrt{25+9})\ \mathrm{m/s^2}} = \frac{34\ \mathrm{N}}{34\ \mathrm{m/s^2}} = 1.00\ \mathrm{kg} \]

Since the object starts from rest, we can use the formula for calculating an object's speed after a given time when the initial velocity is zero: \[ v = at \] Using the magnitude of the acceleration (4.23 m/s²) and the given time (5.00 s): \[ v = (4.23\ \mathrm{m/s^2})(5.00\ \mathrm{s}) = 21.15\ \mathrm{m/s} \]The object's speed after 5.00 s is 21.15 m/s.

To find the components of the velocity vector after 5.00 s, we can multiply the acceleration vector we found in step 3 by the time: \[ \overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{a}}t = (5.00\hat{\mathbf{i}} + 3.00\hat{\mathbf{j}})(5.00\ \mathrm{s}) = 25.00\hat{\mathbf{i}} + 15.00\hat{\mathbf{j}}\ \mathrm{m/s} \] The components of the velocity vector after 5.00 s are \(25.00\hat{\mathbf{i}} + 15.00\hat{\mathbf{j}}\) m/s.