4. A 3.00-kg object has an initial velocity $(6.00\hat{i} - 2.00\hat{j})$ m/s. (a) What is its kinetic energy at this time? (b) Find the total work done on the object as its velocity changes to $(8.00\hat{i} + 4.00\hat{j})$ m/s. (Note: From the definition of the scalar product, $v^2 = \vec{v} \cdot \vec{v}$.) (5 points)
Added by Richard B.
Close
Step 1
Mass of the object, $m = 3.00$ kg. Initial velocity, $\vec{v_i} = (6.00\hat{i} - 2.00\hat{j})$ m/s. Final velocity, $\vec{v_f} = (8.00\hat{i} + 4.00\hat{j})$ m/s. Show more…
Show all steps
Your feedback will help us improve your experience
Nishant Kumar and 96 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A 3.00 -kg object has an initial velocity $(6.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. (a) What is its kinetic energy at this time? (b) Find the total work done on the object as its velocity changes to $(8.00 \hat{\mathbf{i}}+4.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ (Note: From the definition of the scalar product, $\left.v^{2}=\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{v}} .\right)$
Nishant K.
A 3.00 -kg object has a velocity $(6.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}) \quad \mathrm{m} / \mathrm{s}.$ (a) What is its kinetic energy at this time? (b) Find the total work done on the object if its velocity changes to $(8.00 \hat{\mathbf{i}}+4.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ (Note: From the definition of the dot product, $\left.v^{2}=\mathbf{v} \cdot \mathbf{v} .\right)$
A 3.00 -kg object has a velocity $(6.00 \hat{\mathbf{i}}-1.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. (a) What is its kinetic energy at this moment? (b) What is the net work done on the object if its velocity changes to $(8.00 \hat{\mathbf{i}}+4.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ (Note: From the definition of the dot product, $\left.v^{2}=\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{v}} .\right)$
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD