
Concept explainers
What can you say about the velocity of a moving body that in dynamic equilibrium? Draw a sketch of such a body using clearly labeled arrows to represent all external forces on the body.

The velocity of a moving body that is dynamic equilibrium and sketch a diagram of such a body.
Answer to Problem 1CQ
The velocity of a body in dynamic equilibrium is a constant, both in magnitude and direction.
Explanation of Solution
Introduction:
A body is said to be in equilibrium if the total force on the body is zero. On this basis, there are two kinds of equilibria- static equilibrium and dynamic equilibrium. According to Newton's first law, a body continues to be in the state of rest or in the state of uniform motion in a straight line, unless acted upon by an external unbalanced force. Further, Newton's second law implies that an unbalanced force produces an acceleration.
The sum of the forces acting on a body which is in dynamic equilibrium is zero. Therefore, the body experiences no acceleration. As a result, its velocity does not change. Thus, the body which is in dynamic equilibrium moves with a constant velocity.
Consider a block of weight W, acted upon by a force F. The block is placed on a horizontal surface. The surface exerts an upward force FNcalled the normal force on the block. A force of friction f acts between the surface and the block. The block moves towards the right with a velocity v.
The forces W and FNare equal and opposite. Thus, the sum of the forces along the vertical direction is zero. If the applied force F and the force of friction f have equal magnitudes but they are directed opposite to each other. Thus, the net force along the horizontal is also equal to zero. The sum of the forces acting on the block, being zero, the block is not accelerated. Thus, the velocity of the block remains constant. It continues to move with the same speed v and in the same direction.
A diagram representing the forces is shown below.
Conclusion:
A moving body under dynamic equilibrium moves with a constant velocity.
Want to see more full solutions like this?
Chapter 9 Solutions
College Physics
Additional Science Textbook Solutions
Biology: Life on Earth with Physiology (11th Edition)
Applications and Investigations in Earth Science (9th Edition)
Human Anatomy & Physiology (2nd Edition)
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
Campbell Essential Biology (7th Edition)
College Physics: A Strategic Approach (3rd Edition)
- Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 1.00 μC, and L = 0.850 m). Calculate the total electric force on the 7.00-μC charge. magnitude direction N ° (counterclockwise from the +x axis) y 7.00 με 9 L 60.0° x -4.00 μC ①arrow_forward(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 9.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol. (b) Imagine adding electrons to the pin until the negative charge has the very large value 1.00 mC. How many electrons are added for every 109 electrons already present?arrow_forward(a) A physics lab instructor is working on a new demonstration. She attaches two identical copper spheres with mass m = 0.180 g to cords of length L as shown in the figure. A Both spheres have the same charge of 6.80 nC, and are in static equilibrium when 0 = 4.95°. What is L (in m)? Assume the cords are massless. 0.180 Draw a free-body diagram, apply Newton's second law for a particle in equilibrium to one of the spheres. Find an equation for the distance between the two spheres in terms of L and 0, and use this expression in your Coulomb force equation. m (b) What If? The charge on both spheres is increased until each cord makes an angle of 0 = 9.90° with the vertical. If both spheres have the same electric charge, what is the charge (in nC) on each sphere in this case? 9.60 Use the same reasoning as in part (a), only now, use the length found in part (a) and the new angle to solve for the charge. ncarrow_forward
- A proton moves at 5.20 x 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 × 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 6.00 cm horizontally. 83.33 Your response differs from the correct answer by more than 10%. Double check your calculations. ns (b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.) 2.77 Your response differs from the correct answer by more than 10%. Double check your calculations. mm (c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally. = 5.4e5 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. I + 6.68e4 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step…arrow_forward(a) A physics lab instructor is working on a new demonstration. She attaches two identical copper spheres with mass m = 0.180 g to cords of length L as shown in the figure. A Both spheres have the same charge of 6.80 nC, and are in static equilibrium when = 4.95°. What is L (in m)? Assume the cords are massless. 0.150 Draw a free-body diagram, apply Newton's second law for a particle in equilibrium to one of the spheres. Find an equation for the distance between the two spheres in terms of L and 0, and use this expression in your Coulomb force equation. m (b) What If? The charge on both spheres is increased until each cord makes an angle of 0 = 9.90° with the vertical. If both spheres have the same electric charge, what is the charge (in nC) on each sphere in this case? 13.6 ☑ Use the same reasoning as in part (a), only now, use the length found in part (a) and the new angle to solve for the charge. nCarrow_forwardA proton moves at 5.20 x 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 × 10³ N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 6.00 cm horizontally. 1.15e-7 ☑ Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. ns (b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.) 5.33e-3 ☑ Your response is off by a multiple of ten. mm (c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally. | ↑ + jkm/sarrow_forward
- A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 6.00 cm horizontally. (b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.)arrow_forwardThe figure below shows the electric field lines for two charged particles separated by a small distance. 92 91 (a) Determine the ratio 91/92. 1/3 × This is the correct magnitude for the ratio. (b) What are the signs of q₁ and 92? 91 positive 92 negative ×arrow_forwardPlease help me solve this one more detail, thanksarrow_forward
- A dielectric-filled parallel-plate capacitor has plate area A = 20.0 ccm2 , plate separaton d = 10.0 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 . Find the energy U1 of the dielectric-filled capacitor. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?arrow_forwardIn (Figure 1) C1 = 6.00 μF, C2 = 6.00 μF, C3 = 12.0 μF, and C4 = 3.00 μF. The capacitor network is connected to an applied potential difference Vab. After the charges on the capacitors have reached their final values, the voltage across C3 is 40.0 V. What is the voltage across C4? What is the voltage Vab applied to the network? Please explain everything in steps.arrow_forwardI need help with these questions again. A step by step working out with diagrams that explains more clearlyarrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning





