"Vector, NLM And Work Power Energy Online MCQ Test "
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1. If \( \mathbf{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{B} = 3\hat{i} + 2\hat{j} + \hat{k} \), what is the area of the parallelogram formed by \( \mathbf{A} \) and \( \mathbf{B} \)?
A. \( 3 \)
B. \( 8\sqrt{3} \)
C. \( 64 \)
D. \( 0 \)
The correct answer is B: \( 8\sqrt{3} \).
Step-by-Step Solution:
The area of a parallelogram formed by two vectors is given by the magnitude of their cross product: \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(A_y B_z - A_z B_y)^2 + (A_z B_x - A_x B_z)^2 + (A_x B_y - A_y B_x)^2} \] Substituting \( \mathbf{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{B} = 3\hat{i} + 2\hat{j} + \hat{k} \), the cross product and its magnitude yield \( 8\sqrt{3} \).
Step-by-Step Solution:
The area of a parallelogram formed by two vectors is given by the magnitude of their cross product: \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(A_y B_z - A_z B_y)^2 + (A_z B_x - A_x B_z)^2 + (A_x B_y - A_y B_x)^2} \] Substituting \( \mathbf{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{B} = 3\hat{i} + 2\hat{j} + \hat{k} \), the cross product and its magnitude yield \( 8\sqrt{3} \).
2. Two forces of magnitudes \( P \) and \( \sqrt{3P} \) act at right angles to each other. Their resultant makes an angle with \( P \). What is the value of this angle?
A. \( 30^\circ \)
B. \( 45^\circ \)
C. \( 60^\circ \)
D. None of these
The correct answer is C: \( 60^\circ \).
Solution:
The resultant force \( R \) is given by: \[ R = \sqrt{P^2 + (\sqrt{3}P)^2} = \sqrt{4P^2} = 2P \] The angle \( \theta \) is determined using: \[ \tan\theta = \frac{\sqrt{3}P}{P} = \sqrt{3} \quad \Rightarrow \quad \theta = \tan^{-1}(\sqrt{3}) \approx 60^\circ. \]
Solution:
The resultant force \( R \) is given by: \[ R = \sqrt{P^2 + (\sqrt{3}P)^2} = \sqrt{4P^2} = 2P \] The angle \( \theta \) is determined using: \[ \tan\theta = \frac{\sqrt{3}P}{P} = \sqrt{3} \quad \Rightarrow \quad \theta = \tan^{-1}(\sqrt{3}) \approx 60^\circ. \]
3. If the angle between two vectors is \( 60^\circ \), what is the value of \( \mathbf{A} \cdot \mathbf{B} \)?
A. \( \frac{1}{2} |\mathbf{A}| |\mathbf{B}| \)
B. \( 3 |\mathbf{A}| |\mathbf{B}| \)
C. \( |\mathbf{A}| |\mathbf{B}| \)
D. Cannot be determined
The correct answer is A: \( \frac{1}{2} |\mathbf{A}| |\mathbf{B}| \).
Step-by-Step Solution:
1. Dot Product Formula:
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta, \] where \( \theta \) is the angle between the vectors. 2. Substituting the Angle:
Given that the angle between the vectors is \( 60^\circ \), we can substitute into the equation: \[ \cos 60^\circ = \frac{1}{2}. \] So, \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cdot \frac{1}{2}. \] 3. Final Answer:
The answer should be \( \frac{1}{2} |\mathbf{A}| |\mathbf{B}| \).
Step-by-Step Solution:
1. Dot Product Formula:
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta, \] where \( \theta \) is the angle between the vectors. 2. Substituting the Angle:
Given that the angle between the vectors is \( 60^\circ \), we can substitute into the equation: \[ \cos 60^\circ = \frac{1}{2}. \] So, \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cdot \frac{1}{2}. \] 3. Final Answer:
The answer should be \( \frac{1}{2} |\mathbf{A}| |\mathbf{B}| \).
4. Conclusion:
4. A block \( A \) of mass \( 10 \, \text{kg} \) rests on a horizontal surface. The coefficient of friction between block \( A \) and the surface is \( \mu = 0.20 \). A string connects \( A \) to a hanging block \( B \) of unknown mass. The system is set up as shown, with the pulley being frictionless. What is the minimum mass of block \( B \) required to start motion?
A. \( 2 \, \text{kg} \)
B. \( 0.2 \, \text{kg} \)
C. \( 5 \, \text{kg} \)
D. \( 10 \, \text{kg} \)
The correct answer is A: \( 2 \, \text{kg} \).
Step-by-Step Solution:
1. Forces Acting on the System:
- Block \( A \): Frictional force opposes motion. Its maximum value is: \[ f_{\text{max}} = \mu \cdot N = \mu \cdot (m_A g) = 0.20 \cdot (10 \cdot 10) = 20 \, \text{N}. \] - Block \( B \): The downward force due to gravity is: \[ F_B = m_B g. \] 2. Condition for Motion:
For the system to start moving, the force exerted by block \( B \) must exceed the maximum static friction opposing block \( A \): \[ m_B g \geq f_{\text{max}}. \] 3. Solving for \( m_B \):
Substituting values: \[ m_B \cdot 10 \geq 20 \quad \Rightarrow \quad m_B \geq 2 \, \text{kg}. \] 4. Conclusion:
The minimum mass of block \( B \) required to start motion is \( 2 \, \text{kg} \).
Step-by-Step Solution:
1. Forces Acting on the System:
- Block \( A \): Frictional force opposes motion. Its maximum value is: \[ f_{\text{max}} = \mu \cdot N = \mu \cdot (m_A g) = 0.20 \cdot (10 \cdot 10) = 20 \, \text{N}. \] - Block \( B \): The downward force due to gravity is: \[ F_B = m_B g. \] 2. Condition for Motion:
For the system to start moving, the force exerted by block \( B \) must exceed the maximum static friction opposing block \( A \): \[ m_B g \geq f_{\text{max}}. \] 3. Solving for \( m_B \):
Substituting values: \[ m_B \cdot 10 \geq 20 \quad \Rightarrow \quad m_B \geq 2 \, \text{kg}. \] 4. Conclusion:
The minimum mass of block \( B \) required to start motion is \( 2 \, \text{kg} \).
5. A block \( A \) is placed on a rough inclined plane. The coefficient of kinetic friction between the block and the plane is \( \mu_k = \frac{1}{\sqrt{3}} \). The inclination of the plane is gradually increased. When the angle of inclination reaches \( 60^\circ \), the behavior of the block is:
A. Slides down with constant velocity
B. Does not slide
C. Slides down with constant non-zero acceleration
D. Slides down with increasing acceleration
The correct answer is B: Does not slide.
Step-by-Step Solution:
1. Forces Acting on the Block:
- Gravitational force component along the incline: \[ F_{\text{gravity}} = m g \sin\theta. \] - Frictional force opposing motion: \[ F_{\text{friction}} = \mu_k m g \cos\theta. \] 2. Condition for Sliding:
The block does not slide if the frictional force is sufficient to counteract the gravitational component along the incline: \[ F_{\text{friction}} \geq F_{\text{gravity}}. \] 3. Substituting Values:
At \( \theta = 60^\circ \): \[ F_{\text{gravity}} = m g \sin(60^\circ) = m g \cdot \frac{\sqrt{3}}{2}, \] \[ F_{\text{friction}} = \mu_k m g \cos(60^\circ) = {\sqrt{3}} \cdot m g \cdot \frac{1}{2}. \] 4. Comparing Forces:
The ratio of the forces is: \[ \frac{F_{\text{friction}}}{F_{\text{gravity}}} = \frac{\sqrt{3} \cdot m g \cdot \frac{1}{2}}{m g \cdot \frac{\sqrt{3}}{2}} = 1. \] Since \( F_{\text{friction}} = F_{\text{gravity}} \), the block remains stationary.
Step-by-Step Solution:
1. Forces Acting on the Block:
- Gravitational force component along the incline: \[ F_{\text{gravity}} = m g \sin\theta. \] - Frictional force opposing motion: \[ F_{\text{friction}} = \mu_k m g \cos\theta. \] 2. Condition for Sliding:
The block does not slide if the frictional force is sufficient to counteract the gravitational component along the incline: \[ F_{\text{friction}} \geq F_{\text{gravity}}. \] 3. Substituting Values:
At \( \theta = 60^\circ \): \[ F_{\text{gravity}} = m g \sin(60^\circ) = m g \cdot \frac{\sqrt{3}}{2}, \] \[ F_{\text{friction}} = \mu_k m g \cos(60^\circ) = {\sqrt{3}} \cdot m g \cdot \frac{1}{2}. \] 4. Comparing Forces:
The ratio of the forces is: \[ \frac{F_{\text{friction}}}{F_{\text{gravity}}} = \frac{\sqrt{3} \cdot m g \cdot \frac{1}{2}}{m g \cdot \frac{\sqrt{3}}{2}} = 1. \] Since \( F_{\text{friction}} = F_{\text{gravity}} \), the block remains stationary.
5. Conclusion:
6. The momentum of a body is increased by \( 50\% \). By what percentage will its kinetic energy increase?
A. \( 50\% \)
B. \( 125\% \)
C. \( 330\% \)
D. \( 400\% \)
The correct answer is B: \( 125\% \).
Step-by-Step Solution:
1. Relation Between Kinetic Energy and Momentum:
Kinetic energy \( KE \) is proportional to the square of momentum \( p \): \[ KE = \frac{p^2}{2m}. \] 2. Change in Momentum:
If \( p \) increases by \( 50\% \), the new momentum is: \[ p_{\text{new}} = 1.5p. \] 3. Change in Kinetic Energy:
Substituting \( p_{\text{new}} \) into the kinetic energy formula: \[ KE_{\text{new}} = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m}. \] Thus: \[ KE_{\text{new}} = 2.25 \cdot KE_{\text{old}}. \] 4. Percentage Increase:
The increase in kinetic energy is: \[ \Delta KE = 2.25 \cdot KE - KE = 1.25 \cdot KE. \] Therefore, the kinetic energy increases by \( 125\% \).
Step-by-Step Solution:
1. Relation Between Kinetic Energy and Momentum:
Kinetic energy \( KE \) is proportional to the square of momentum \( p \): \[ KE = \frac{p^2}{2m}. \] 2. Change in Momentum:
If \( p \) increases by \( 50\% \), the new momentum is: \[ p_{\text{new}} = 1.5p. \] 3. Change in Kinetic Energy:
Substituting \( p_{\text{new}} \) into the kinetic energy formula: \[ KE_{\text{new}} = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m}. \] Thus: \[ KE_{\text{new}} = 2.25 \cdot KE_{\text{old}}. \] 4. Percentage Increase:
The increase in kinetic energy is: \[ \Delta KE = 2.25 \cdot KE - KE = 1.25 \cdot KE. \] Therefore, the kinetic energy increases by \( 125\% \).
7. A block of mass \( 10 \, \text{kg} \) is released on a fixed wedge inside a cart moving with a constant velocity of \( 10 \, \text{m/s} \). The initial velocity of the block with respect to the cart is zero. What is the work done by the normal reaction (with respect to the ground) on the block in \( 2 \, \text{s} \)? \( g = 10 \, \text{m/s}^2 \).
A. \( 0 \, \text{J} \)
B. \( 960 \, \text{J} \)
C. \( 1200 \, \text{J} \)
D. None of these
The correct answer is B: \( 960 \, \text{J} \).
Step-by-Step Solution:
1. Key Forces and Displacements:
- The acceleration of the wedge is \( 0 \), so the normal reaction \( N \) is: \[ N = m g \cos \theta. \] - The displacement of the block normal to the incline in \( t = 2 \, \text{s} \) is: \[ S = v \sin \theta \cdot t = 10 \cdot \frac{3}{5} \cdot 2 = 12 \, \text{m}. \] 2. Work Done by Normal Reaction:
\[ W = N \cdot S = m g \cos \theta \cdot S = 100 \cdot \frac{4}{5} \cdot 12 = 960 \, \text{J}. \] 3. Conclusion:
The work done by the normal reaction is \( 960 \, \text{J} \).
Step-by-Step Solution:
1. Key Forces and Displacements:
- The acceleration of the wedge is \( 0 \), so the normal reaction \( N \) is: \[ N = m g \cos \theta. \] - The displacement of the block normal to the incline in \( t = 2 \, \text{s} \) is: \[ S = v \sin \theta \cdot t = 10 \cdot \frac{3}{5} \cdot 2 = 12 \, \text{m}. \] 2. Work Done by Normal Reaction:
\[ W = N \cdot S = m g \cos \theta \cdot S = 100 \cdot \frac{4}{5} \cdot 12 = 960 \, \text{J}. \] 3. Conclusion:
The work done by the normal reaction is \( 960 \, \text{J} \).
8. A block of mass \( m \) starts at rest at a height \( h \) on a frictionless inclined plane. It slides down, travels across a rough horizontal surface with coefficient of kinetic friction \( \mu_k \), and compresses a spring with constant \( k \) a distance \( x \). The block momentarily comes to rest and then travels back up. What is the maximum height \( h' \) reached by the block upon return?
A. \( m g h' = m g h - \mu_k m g d \)
B. \( m g h' = m g h + m g d \)
C. \( m g h' = m g h + m g d + k x^2 \)
D. \( m g h' = m g h - m g d - k x^2 \)
The correct answer is A: \( m g h' = m g h - m g d \).
Step-by-Step Solution:
1. Energy Conservation:
- Initial potential energy at height \( h \): \[ U = m g h. \] - Work done against friction over distance \( d \): \[ W_{\text{friction}} = \mu_k m g d. \] - Final potential energy at height \( h' \): \[ U' = m g h'. \] 2. Energy Balance:
Total energy at \( h' \) is reduced by the work done: \[ m g h' = m g h - \mu_k m g d. \] 3. Simplification:
Factoring out \( m g \): \[ h' = h - \mu_k d. \] 4. Conclusion:
The maximum height reached by the block is \( m g h' = m g h - \mu_k m g d \).
Step-by-Step Solution:
1. Energy Conservation:
- Initial potential energy at height \( h \): \[ U = m g h. \] - Work done against friction over distance \( d \): \[ W_{\text{friction}} = \mu_k m g d. \] - Final potential energy at height \( h' \): \[ U' = m g h'. \] 2. Energy Balance:
Total energy at \( h' \) is reduced by the work done: \[ m g h' = m g h - \mu_k m g d. \] 3. Simplification:
Factoring out \( m g \): \[ h' = h - \mu_k d. \] 4. Conclusion:
The maximum height reached by the block is \( m g h' = m g h - \mu_k m g d \).
9. A block of mass \( 5 \, \text{kg} \) is released from rest when a spring is compressed by \( 2 \, \text{m} \). The block is not attached to the spring, and the spring's natural length is \( 4 \, \text{m} \). What is the maximum height reached by the block from the ground? \( g = 10 \, \text{m/s}^2 \).
A. \( 5.5 \, \text{m} \)
B. \( 4.5 \, \text{m} \)
C. \( 6 \, \text{m} \)
D. \( 7.5 \, \text{m} \)
The correct answer is A: \( 5.5 \, \text{m} \).
Step-by-Step Solution:
1. Conservation of Momentum:
3. Conclusion:
The maximum height from the ground is \( 5.5 \, \text{m} \).
Step-by-Step Solution:
1. Conservation of Momentum:
The maximum height from the ground is \( 5.5 \, \text{m} \).
10. A particle moves along a circle of radius \( R \) with constant angular speed \( \omega \). What is the displacement (only magnitude) of the particle in time \( t \)?
A. \( R t \)
B. \( 2R \cos(\omega t) \)
C. \( 2R \sin(\omega t) \)
D. \( 2R \sin\left(\frac{\omega t}{2}\right) \)
The correct answer is D: \( 2R \sin\left(\frac{\omega t}{2}\right) \).
4. Conclusion:
The displacement is \( 2R \sin\left(\frac{\omega t}{2}\right) \).
The displacement is \( 2R \sin\left(\frac{\omega t}{2}\right) \).
11. A table fan rotating at \( 2400 \, \text{rpm} \) is switched off, and the rpm decreases linearly with time until the fan comes to rest in \( 8 \, \text{s} \). What is the total number of revolutions made by the fan before coming to rest?
A. \( 420 \)
B. \( 280 \)
C. \( 190 \)
D. \( 16800 \)
The correct answer is B: \( 280 \).
4. Conclusion:
The fan completes \( 280 \) revolutions before coming to rest.
The fan completes \( 280 \) revolutions before coming to rest.
12. A body of mass \( m \) is moving in a circle of radius \( r \) with a constant speed \( v \). The force on the body is \( F = \frac{mv^2}{r} \), directed towards the center. What is the work done by this force in moving the body halfway around the circle?
A. \( \frac{mv^2}{r} \)
B. \( \frac{1}{2} mv^2 \)
C. \( mv^2 \)
D. \( 0 \)
The correct answer is D: \( 0 \).
Step-by-Step Solution:
1. Work Done by a Force:
Work is defined as: \[ W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos\theta, \] where \( \theta \) is the angle between \( \vec{F} \) and \( \vec{d} \). 2. Force in Circular Motion:
In uniform circular motion: - The centripetal force \( \vec{F} \) always points towards the center. - The displacement \( \vec{d} \) is tangential to the circle. - Thus, \( \theta = 90^\circ \), and \( \cos(90^\circ) = 0 \). 3. Work Done:
Since \( \cos\theta = 0 \): \[ W = |\vec{F}| |\vec{d}| \cos\theta = 0. \] 4. Conclusion:
The work done by the centripetal force is \( 0 \).
Step-by-Step Solution:
1. Work Done by a Force:
Work is defined as: \[ W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos\theta, \] where \( \theta \) is the angle between \( \vec{F} \) and \( \vec{d} \). 2. Force in Circular Motion:
In uniform circular motion: - The centripetal force \( \vec{F} \) always points towards the center. - The displacement \( \vec{d} \) is tangential to the circle. - Thus, \( \theta = 90^\circ \), and \( \cos(90^\circ) = 0 \). 3. Work Done:
Since \( \cos\theta = 0 \): \[ W = |\vec{F}| |\vec{d}| \cos\theta = 0. \] 4. Conclusion:
The work done by the centripetal force is \( 0 \).
13. Which of the following is not a unit of length?
A. Micron
B. Light year
C. Angstrom
D. Radian
The correct answer is D: Radian.
Step-by-Step Solution:
1. Understanding Units:
- A *micron* (\( 10^{-6} \, \text{m} \)), *light year* (\( 9.46 \times 10^{15} \, \text{m} \)), and *angstrom* (\( 10^{-10} \, \text{m} \)) are all units of length.
- A *radian* is a unit of angular measurement, not length.
2. Conclusion:
Since a radian is not a unit of length, it is the correct answer.
Step-by-Step Solution:
1. Understanding Units:
- A *micron* (\( 10^{-6} \, \text{m} \)), *light year* (\( 9.46 \times 10^{15} \, \text{m} \)), and *angstrom* (\( 10^{-10} \, \text{m} \)) are all units of length.
- A *radian* is a unit of angular measurement, not length.
2. Conclusion:
Since a radian is not a unit of length, it is the correct answer.
14. A stone of mass \( 0.5 \, \text{kg} \) tied to a string of length \( 1 \, \text{m} \) moves in a horizontal circular path with a speed of \( 4 \, \text{m/s} \). What is the tension in the string in newtons?
A. \( 2 \, \text{N} \)
B. \( 8 \, \text{N} \)
C. \( 0.2 \, \text{N} \)
D. \( 0.8 \, \text{N} \)
The correct answer is B: \( 8 \, \text{N} \).
Step-by-Step Solution:
1. Centripetal Force Formula:
The tension in the string provides the centripetal force required for circular motion: \[ T = \frac{m v^2}{r}. \] 2. Substituting Values:
Given \( m = 0.5 \, \text{kg} \), \( v = 4 \, \text{m/s} \), and \( r = 1 \, \text{m} \): \[ T = \frac{0.5 \cdot 4^2}{1} = \frac{0.5 \cdot 16}{1} = 8 \, \text{N}. \] 3. Conclusion:
The tension in the string is \( 8 \, \text{N} \).
Step-by-Step Solution:
1. Centripetal Force Formula:
The tension in the string provides the centripetal force required for circular motion: \[ T = \frac{m v^2}{r}. \] 2. Substituting Values:
Given \( m = 0.5 \, \text{kg} \), \( v = 4 \, \text{m/s} \), and \( r = 1 \, \text{m} \): \[ T = \frac{0.5 \cdot 4^2}{1} = \frac{0.5 \cdot 16}{1} = 8 \, \text{N}. \] 3. Conclusion:
The tension in the string is \( 8 \, \text{N} \).
15. A heavy mass is attached to a thin wire and is whirled in a vertical circle. At which point is the wire most likely to break?
A. At the highest point of the circle
B. At the lowest point of the circle
C. When the wire is horizontal
D. At an angle of \( \cos^{-1}(1/3) \) from the upward vertical
The correct answer is B: At the lowest point of the circle.
Step-by-Step Solution:
1. Forces in Vertical Circular Motion:
- At the highest point, tension \( T \) is minimum because gravity adds to the centripetal force requirement: \[ T = \frac{m v^2}{r} - m g. \] - At the lowest point, tension \( T \) is maximum because gravity opposes the centripetal force: \[ T = \frac{m v^2}{r} + m g. \] 2. Likelihood of Wire Breaking:
The wire is most likely to break at the lowest point where the tension is highest. 3. Conclusion:
The wire will break at the lowest point of the circle.
Step-by-Step Solution:
1. Forces in Vertical Circular Motion:
- At the highest point, tension \( T \) is minimum because gravity adds to the centripetal force requirement: \[ T = \frac{m v^2}{r} - m g. \] - At the lowest point, tension \( T \) is maximum because gravity opposes the centripetal force: \[ T = \frac{m v^2}{r} + m g. \] 2. Likelihood of Wire Breaking:
The wire is most likely to break at the lowest point where the tension is highest. 3. Conclusion:
The wire will break at the lowest point of the circle.
16. An engine exerts a force \( \vec{F} = -20\hat{i} + 3\hat{j} + 5\hat{k} \, \text{N} \) and moves with a velocity \( \vec{v} = 6\hat{i} - 20\hat{j} - 3\hat{k} \, \text{m/s} \). What is the power of the engine (in watts)?
A. \( 45 \, \text{W} \)
B. \( 75 \, \text{W} \)
C. \( 20 \, \text{W} \)
D. \( 10 \, \text{W} \)
The correct answer is A: \( 45 \, \text{W} \).
Step-by-Step Solution:
1. Power Formula:
Power is the scalar product of force and velocity: \[ P = \vec{F} \cdot \vec{v}. \] 2. Dot Product Calculation:
Substituting \( \vec{F} = -20\hat{i} + 3\hat{j} + 5\hat{k} \) and \( \vec{v} = 6\hat{i} - 20\hat{j} - 3\hat{k} \): \[ P = (-20)(6) + (3)(-20) + (5)(-3). \] Simplifying: \[ P = 120 - 60 - 15 = 45 \, \text{W}. \] 3. Power Value:
The magnitude is \( 45 \, \text{W} \), with direction indicating the negative work done due to opposing motion. Conclusion:
The engine's power output is \( 45 \, \text{W} \).
Step-by-Step Solution:
1. Power Formula:
Power is the scalar product of force and velocity: \[ P = \vec{F} \cdot \vec{v}. \] 2. Dot Product Calculation:
Substituting \( \vec{F} = -20\hat{i} + 3\hat{j} + 5\hat{k} \) and \( \vec{v} = 6\hat{i} - 20\hat{j} - 3\hat{k} \): \[ P = (-20)(6) + (3)(-20) + (5)(-3). \] Simplifying: \[ P = 120 - 60 - 15 = 45 \, \text{W}. \] 3. Power Value:
The magnitude is \( 45 \, \text{W} \), with direction indicating the negative work done due to opposing motion. Conclusion:
The engine's power output is \( 45 \, \text{W} \).
17. What is the formula for centripetal acceleration in a circular motion?
A. \( a_c = r \)
B. \( a_c = \frac{v^2}{r} \)
C. \( a_c = v \)
D. \( a_c = r v^2 \)
The correct answer is B: \( a_c = \frac{v^2}{r} \).
Step-by-Step Solution:
1. Centripetal Acceleration Definition:
Centripetal acceleration is the inward acceleration required to keep an object moving in a circular path at a constant speed.
2. Derivation:
- For a body moving in a circular path of radius \( r \) with speed \( v \), the centripetal acceleration is: \[ a_c = \frac{v^2}{r}. \] 3. Common Misconceptions:
- \( a_c = v \): Incorrect, as acceleration depends on \( v^2 \).
- \( a_c = r v^2 \): Incorrect, as radius appears in the denominator. 4. Conclusion:
The correct formula is \( a_c = \frac{v^2}{r} \).
Step-by-Step Solution:
1. Centripetal Acceleration Definition:
Centripetal acceleration is the inward acceleration required to keep an object moving in a circular path at a constant speed.
2. Derivation:
- For a body moving in a circular path of radius \( r \) with speed \( v \), the centripetal acceleration is: \[ a_c = \frac{v^2}{r}. \] 3. Common Misconceptions:
- \( a_c = v \): Incorrect, as acceleration depends on \( v^2 \).
- \( a_c = r v^2 \): Incorrect, as radius appears in the denominator. 4. Conclusion:
The correct formula is \( a_c = \frac{v^2}{r} \).
18. A vector \( \mathbf{A} = 6\hat{i} - 3\hat{j} + 2\hat{k} \) is given. What is the component of \( \mathbf{A} \) along \( \mathbf{B} = 2\hat{i} + 2\hat{j} + \hat{k} \)?
A. \( \frac{12}{3} \)
B. \( \frac{10}{3} \)
C. \( \frac{8}{3} \)
D. \( \frac{9}{3} \)
The correct answer is C: \( \frac{8}{3} \).
Step-by-Step Solution:
1. Formula:
The component of \( \mathbf{A} \) along \( \mathbf{B} \) is: \[ \text{Component} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}. \] 2. Dot Product:
\[ \mathbf{A} \cdot \mathbf{B} = (6)(2) + (-3)(2) + (2)(1) = 12 - 6 + 2 = 8. \] 3. Magnitude of \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3. \] 4. Component Calculation:
\[ \text{Component} = \frac{8}{3}. \] Conclusion:
The component of \( \mathbf{A} \) along \( \mathbf{B} \) is \( \frac{8}{3} \).
Step-by-Step Solution:
1. Formula:
The component of \( \mathbf{A} \) along \( \mathbf{B} \) is: \[ \text{Component} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}. \] 2. Dot Product:
\[ \mathbf{A} \cdot \mathbf{B} = (6)(2) + (-3)(2) + (2)(1) = 12 - 6 + 2 = 8. \] 3. Magnitude of \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3. \] 4. Component Calculation:
\[ \text{Component} = \frac{8}{3}. \] Conclusion:
The component of \( \mathbf{A} \) along \( \mathbf{B} \) is \( \frac{8}{3} \).
19. Two vectors \( \mathbf{A} = 3\hat{i} + 4\hat{j} \) and \( \mathbf{B} = 5\hat{i} + 12\hat{j} \) are given. What is the angle \( \theta \) between \( \mathbf{A} \) and \( \mathbf{B} \)?
A. \( 30^\circ \)
B. \( 45^\circ \)
C. \( \cos^{-1}\left(\frac{69}{80}\right) \)
D. \( 60^\circ \)
The correct answer is C: \( \cos^{-1}\left(\frac{69}{80}\right) \).
Step-by-Step Solution:
1. Dot Product Formula:
The dot product of two vectors is given by: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta. \] 2. Calculate \( \mathbf{A} \cdot \mathbf{B} \):
\[ \mathbf{A} \cdot \mathbf{B} = (3)(5) + (4)(12) = 15 + 48 = 63. \] 3. Magnitudes of Vectors:
\[ |\mathbf{A}| = \sqrt{3^2 + 4^2} = 5, \quad |\mathbf{B}| = \sqrt{5^2 + 12^2} = 13. \] 4. Substituting Values:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{63}{5 \cdot 13} = \frac{69}{80}. \] Thus, \[ \theta = \cos^{-1}\left(\frac{69}{80}\right). \] Conclusion:
The angle between the vectors is \( \cos^{-1}\left(\frac{69}{80}\right) \).
Step-by-Step Solution:
1. Dot Product Formula:
The dot product of two vectors is given by: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta. \] 2. Calculate \( \mathbf{A} \cdot \mathbf{B} \):
\[ \mathbf{A} \cdot \mathbf{B} = (3)(5) + (4)(12) = 15 + 48 = 63. \] 3. Magnitudes of Vectors:
\[ |\mathbf{A}| = \sqrt{3^2 + 4^2} = 5, \quad |\mathbf{B}| = \sqrt{5^2 + 12^2} = 13. \] 4. Substituting Values:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{63}{5 \cdot 13} = \frac{69}{80}. \] Thus, \[ \theta = \cos^{-1}\left(\frac{69}{80}\right). \] Conclusion:
The angle between the vectors is \( \cos^{-1}\left(\frac{69}{80}\right) \).
20. Two vectors are given as \( \mathbf{A} = 2\hat{i} + \hat{j} \) and \( \mathbf{B} = \hat{i} - 3\hat{j} + \hat{k} \). What is the unit vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \)?
A. \( \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - 7\hat{k}) \)
B. \( \frac{1}{\sqrt{14}}(-3\hat{i} + 2\hat{j} + 7\hat{k}) \)
C. \( \frac{1}{14}(3\hat{i} - 2\hat{j} - 7\hat{k}) \)
D. \( \frac{1}{14}(-3\hat{i} - 2\hat{j} + 7\hat{k}) \)
The correct answer is A: \( \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - 7\hat{k}) \).
Step-by-Step Solution:
1. Cross Product for Perpendicular Vector:
The vector perpendicular to \( \mathbf{A} \) and \( \mathbf{B} \) is given by \( \mathbf{A} \times \mathbf{B} \): \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 1 & -3 & 1 \end{vmatrix}. \] 2. Determinant Expansion:
Expanding the determinant: \[ \mathbf{A} \times \mathbf{B} = \hat{i}\begin{vmatrix} 1 & 0 \\ -3 & 1 \end{vmatrix} - \hat{j}\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} + \hat{k}\begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix}. \] Simplifying: \[ \mathbf{A} \times \mathbf{B} = \hat{i}(1 \cdot 1 - 0 \cdot -3) - \hat{j}(2 \cdot 1 - 0 \cdot 1) + \hat{k}(2 \cdot -3 - 1 \cdot 1). \] \[ \mathbf{A} \times \mathbf{B} = \hat{i}(1) - \hat{j}(2) + \hat{k}(-6 - 1). \] \[ \mathbf{A} \times \mathbf{B} = \hat{i} + -2\hat{j} - 7\hat{k}. \] 3. Normalize the Vector:
The unit vector is: \[ \mathbf{u} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|}. \] Magnitude: \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{1^2 + (-2)^2 + (-7)^2} = \sqrt{1 + 4 + 49} = \sqrt{14}. \] Unit vector: \[ \mathbf{u} = \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - 7\hat{k}). \] 4. Conclusion:
The unit vector perpendicular to \( \mathbf{A} \) and \( \mathbf{B} \) is \( \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - 7\hat{k}) \).
Step-by-Step Solution:
1. Cross Product for Perpendicular Vector:
The vector perpendicular to \( \mathbf{A} \) and \( \mathbf{B} \) is given by \( \mathbf{A} \times \mathbf{B} \): \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 1 & -3 & 1 \end{vmatrix}. \] 2. Determinant Expansion:
Expanding the determinant: \[ \mathbf{A} \times \mathbf{B} = \hat{i}\begin{vmatrix} 1 & 0 \\ -3 & 1 \end{vmatrix} - \hat{j}\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} + \hat{k}\begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix}. \] Simplifying: \[ \mathbf{A} \times \mathbf{B} = \hat{i}(1 \cdot 1 - 0 \cdot -3) - \hat{j}(2 \cdot 1 - 0 \cdot 1) + \hat{k}(2 \cdot -3 - 1 \cdot 1). \] \[ \mathbf{A} \times \mathbf{B} = \hat{i}(1) - \hat{j}(2) + \hat{k}(-6 - 1). \] \[ \mathbf{A} \times \mathbf{B} = \hat{i} + -2\hat{j} - 7\hat{k}. \] 3. Normalize the Vector:
The unit vector is: \[ \mathbf{u} = \frac{\mathbf{A} \times \mathbf{B}}{|\mathbf{A} \times \mathbf{B}|}. \] Magnitude: \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{1^2 + (-2)^2 + (-7)^2} = \sqrt{1 + 4 + 49} = \sqrt{14}. \] Unit vector: \[ \mathbf{u} = \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - 7\hat{k}). \] 4. Conclusion:
The unit vector perpendicular to \( \mathbf{A} \) and \( \mathbf{B} \) is \( \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - 7\hat{k}) \).
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