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Math  /  Algebra

Question42. (II) A 3.0-kg object has the following two forces acting on it: Fundefined1=(16i^+12j^)NFundefined2=(−10i^+22j^)N\begin{array}{l} \overrightarrow{\mathbf{F}}_{1}=(16 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}) \mathbf{N} \\ \overrightarrow{\mathbf{F}}_{2}=(-10 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}) \mathbf{N} \end{array}
If the object is initially at rest, determine its velocity vundefined\overrightarrow{\mathbf{v}} at t=4.0 st=4.0 \mathrm{~s}.

Studdy Solution

STEP 1

1. The object is subject to Newton's second law of motion.
2. The net force acting on the object is the vector sum of the given forces.
3. The object is initially at rest, meaning its initial velocity is zero.
4. The acceleration is constant over the time interval.

STEP 2

1. Calculate the net force acting on the object.
2. Determine the acceleration of the object.
3. Calculate the velocity of the object at t=4.0 s t = 4.0 \, \text{s} .

STEP 3

Calculate the net force by adding the two given forces vectorially:
Fundefinednet=Fundefined1+Fundefined2\overrightarrow{\mathbf{F}}_{\text{net}} = \overrightarrow{\mathbf{F}}_{1} + \overrightarrow{\mathbf{F}}_{2}
Substitute the given forces:
Fundefinednet=(16i^+12j^) N+(−10i^+22j^) N\overrightarrow{\mathbf{F}}_{\text{net}} = (16 \hat{\mathbf{i}} + 12 \hat{\mathbf{j}}) \, \text{N} + (-10 \hat{\mathbf{i}} + 22 \hat{\mathbf{j}}) \, \text{N}
Combine the components:
Fundefinednet=(16−10)i^+(12+22)j^ N\overrightarrow{\mathbf{F}}_{\text{net}} = (16 - 10) \hat{\mathbf{i}} + (12 + 22) \hat{\mathbf{j}} \, \text{N}
Fundefinednet=6i^+34j^ N\overrightarrow{\mathbf{F}}_{\text{net}} = 6 \hat{\mathbf{i}} + 34 \hat{\mathbf{j}} \, \text{N}

STEP 4

Use Newton's second law to find the acceleration:
Fundefinednet=maundefined\overrightarrow{\mathbf{F}}_{\text{net}} = m \overrightarrow{\mathbf{a}}
Solve for acceleration aundefined\overrightarrow{\mathbf{a}}:
aundefined=Fundefinednetm\overrightarrow{\mathbf{a}} = \frac{\overrightarrow{\mathbf{F}}_{\text{net}}}{m}
Substitute the known values:
aundefined=6i^+34j^3.0 m/s2\overrightarrow{\mathbf{a}} = \frac{6 \hat{\mathbf{i}} + 34 \hat{\mathbf{j}}}{3.0} \, \text{m/s}^2
Calculate the components:
aundefined=2i^+343j^ m/s2\overrightarrow{\mathbf{a}} = 2 \hat{\mathbf{i}} + \frac{34}{3} \hat{\mathbf{j}} \, \text{m/s}^2
aundefined=2i^+11.33j^ m/s2\overrightarrow{\mathbf{a}} = 2 \hat{\mathbf{i}} + 11.33 \hat{\mathbf{j}} \, \text{m/s}^2

STEP 5

Use the kinematic equation for velocity with constant acceleration:
vundefined=vundefined0+aundefinedt\overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{v}}_0 + \overrightarrow{\mathbf{a}} t
Since the object is initially at rest, vundefined0=0\overrightarrow{\mathbf{v}}_0 = 0:
vundefined=aundefinedt\overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{a}} t
Substitute the known values:
vundefined=(2i^+11.33j^) m/s2×4.0 s\overrightarrow{\mathbf{v}} = (2 \hat{\mathbf{i}} + 11.33 \hat{\mathbf{j}}) \, \text{m/s}^2 \times 4.0 \, \text{s}
Calculate the components:
vundefined=(2×4.0)i^+(11.33×4.0)j^ m/s\overrightarrow{\mathbf{v}} = (2 \times 4.0) \hat{\mathbf{i}} + (11.33 \times 4.0) \hat{\mathbf{j}} \, \text{m/s}
vundefined=8i^+45.32j^ m/s\overrightarrow{\mathbf{v}} = 8 \hat{\mathbf{i}} + 45.32 \hat{\mathbf{j}} \, \text{m/s}
The velocity of the object at t=4.0 s t = 4.0 \, \text{s} is:
vundefined=8i^+45.32j^ m/s\overrightarrow{\mathbf{v}} = 8 \hat{\mathbf{i}} + 45.32 \hat{\mathbf{j}} \, \text{m/s}

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