At the instant that the velocity of the crate is (3.80 m/s)i + (2.20 m/s)j, what is the instantaneous power supplied by this force? Express your answer with the appropriate units.
Understand the Problem
The question is asking to calculate the instantaneous power supplied by a force acting on a crate sliding on a factory floor, given the force vector and the velocity vector at a certain moment. The instantaneous power can be determined using the dot product of the force vector and the velocity vector.
Answer
The instantaneous power is $$ -25.7 \, \text{W} $$.
Answer for screen readers
The instantaneous power supplied by the force is $$ -25.7 , \text{W} $$.
Steps to Solve
- Identify the Force and Velocity Vectors
The force vector is given by $$ \vec{F} = (-8.50 , \text{N}) \hat{i} + (3.00 , \text{N}) \hat{j} $$ and the velocity vector is $$ \vec{v} = (3.80 , \text{m/s}) \hat{i} + (2.20 , \text{m/s}) \hat{j} $$.
- Calculate the Dot Product
To find the instantaneous power, calculate the dot product of the force vector and the velocity vector. The formula for the dot product is: $$ \vec{F} \cdot \vec{v} = F_x v_x + F_y v_y $$
Substituting in the components: $$ \vec{F} \cdot \vec{v} = (-8.50 , \text{N}) (3.80 , \text{m/s}) + (3.00 , \text{N}) (2.20 , \text{m/s}) $$
- Compute Each Term of the Dot Product
Calculate the two terms separately:
- First term: $$ (-8.50 \times 3.80) = -32.3 , \text{W} $$
- Second term: $$ (3.00 \times 2.20) = 6.6 , \text{W} $$
- Sum the Results
Now add the two results together: $$ \text{Power} = -32.3 , \text{W} + 6.6 , \text{W} = -25.7 , \text{W} $$
The instantaneous power supplied by the force is $$ -25.7 , \text{W} $$.
More Information
This negative power indicates that the force is opposing the motion of the crate, suggesting it's resisting the crate's movement in the direction of the applied force. Instantaneous power can provide insight into the effectiveness of the force in moving the object.
Tips
- Forgetting to use the correct signs: When dealing with vectors, be careful with the signs of the components. Negative signs indicate direction and must be included in calculations.
- Misapplying the dot product: Ensure that you are performing the dot product accurately by multiplying corresponding components correctly.
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