Derive the formula for the kinetic energy of an object of mass m moving with velocity v

Learning Objectives By the end of this section, you will be able to: • Describe the differences between rotational and translational kinetic energy • Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis • Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy • Use conservation of mechanical energy to analyze systems undergoing both rotation and translation • Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces So far in this chapter, we have been working with rotational kinematics: the description of motion for a rotating rigid body with a fixed axis of rotation. In this section, we define two new quantities that are helpful for analyzing properties of rotating objects: moment of inertia and rotational kinetic energy. With these properties defined, we will have two important tools we need for analyzing rotational dynamics. Rotational Kinetic Energy Any moving object has kinetic energy. We know how to calculate this for a body undergoing translational motion, but how about for a rigid body undergoing rotation? This might seem complicated because each point on the rigid body has a different velocity. However, we can make use of angular velocity—which is the same for the entire rigid body—to express the kinetic energy for a rotating object. rotational kinetic energy. Figure 10.17 The rotational kinetic energy of the grindstone is...

Let the initial velocity of the object be u. Let an external force be applied on it so that gets displaced by distance s and its velocity becomes v. In this scenario, the kinetic energy of the moving body is equal to the work that was required to change its velocity from u to v. thus, we have the velocity- position relation as: V 2 = u 2 + 2 a s or s = 2 a v 2 − u 2 ​ ....(i) Where, a is the acceleration of the during the change in its velocity. Now, the work done on the body by the external force is given by: W = F × s F = ma ... (ii) from equations (i) and (ii), we obtain : W = m a × ( 2 a v 2 − u 2 ​ ) = 2 1 ​ m ( v 2 − u 2 ) If the body was initially at rest 9 i. e, u = 0 ), then : W = 2 1 ​ m v 2 Since kinetic energy is equal to the work done on the body to change its velocity from 0 to v , we obtain: kinetic energy , E k ​ = 2 1 ​ m v 2

What does "best" mean? (one dimensional case) start with relativistic momentum ##p = \gamma(u) m u## Force: ##F_\text## What does "best" mean? (one dimensional case) start with relativistic momentum ##p = \gamma(u) m u## Force: ##F_\text x = mc^2\cdot \gamma (u(x_2)) -mc^2\cdot \gamma(u(x_1)) ## Darn, I'm on my phone, cannot see posted equations. Best? Occam's razor kind of best. E = mc^2 is also not even the full story. It is a special case of something having zero speed as measured in a certain frame of reference. No aether, what was it- splitting light and bouncing back no matter how the thing is rotated with respect to the rotating earth that is flying around the sun. I'm on vacation, just looking for book suggestions. No aether, what was it- splitting light and bouncing back no matter how the thing is rotated with respect to the rotating earth that is flying around the sun. I'm on vacation, just looking for book suggestions. There are plenty of threads where people are asking for book recommendations regarding learning special relativity. I can suggest the book by Morin Is there any relationship between: E = m v^2 and E = Pressure*Volume? PV = (density *Volume) * velocity^2 Pressure=density*velocity^2 almost starts looking like Bernoilli's equation. Of course, for dimensional reasons, one can see resemblances to possibly familiar equations. While these may hint at connections from relativistic objects, the appropriate path is to study the relativistic versions of thos...

The kinetic energy of a body depends upon mass (m) and velocity (v) of the body. Let K.E. ∝ m x v y ∴ K.E. = km x v y .....(1) where, k = dimensionless constant of proportionality. Taking dimensions on both sides of equation (1), [L 2M 1T -2] = [L 0 M 1 T 0] x [L 1 M 0 T -1] y = [L 0 M x T 0] [L y M 0 T -y] = [L (0+y) M (x + 0) T (0- y)] [L 2 M 1 T -2] = [L y M x T -y]....(2) Equating dimensions of L, M, T on both sides of equation (2), x = 1 and y = 2 Substituting x, y in equation (1), we have K.E. = kmv 2

$$ KE = \dfracmv^2 $$ Where: • KE = kinetic energy • m = mass of a body • v = velocity of a body Kinetic Energy Kinetic Energy is the energy an object has owing to its motion. In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2. We use Joules, kilograms, and meters per second as our defaults, although any appropriate units for mass (grams, ounces, etc.) or velocity (miles per hour, millimeters per second, etc.) could certainly be used as well - the calculation is the same regardless. References/ Further Reading • •

Let the initial velocity of the object be u. Let an external force be applied on it so that gets displaced by distance s and its velocity becomes v. In this scenario, the kinetic energy of the moving body is equal to the work that was required to change its velocity from u to v. thus, we have the velocity- position relation as: V 2 = u 2 + 2 a s or s = 2 a v 2 − u 2 ​ ....(i) Where, a is the acceleration of the during the change in its velocity. Now, the work done on the body by the external force is given by: W = F × s F = ma ... (ii) from equations (i) and (ii), we obtain : W = m a × ( 2 a v 2 − u 2 ​ ) = 2 1 ​ m ( v 2 − u 2 ) If the body was initially at rest 9 i. e, u = 0 ), then : W = 2 1 ​ m v 2 Since kinetic energy is equal to the work done on the body to change its velocity from 0 to v , we obtain: kinetic energy , E k ​ = 2 1 ​ m v 2

Learning Objectives By the end of this section, you will be able to: • Describe the differences between rotational and translational kinetic energy • Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis • Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy • Use conservation of mechanical energy to analyze systems undergoing both rotation and translation • Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces So far in this chapter, we have been working with rotational kinematics: the description of motion for a rotating rigid body with a fixed axis of rotation. In this section, we define two new quantities that are helpful for analyzing properties of rotating objects: moment of inertia and rotational kinetic energy. With these properties defined, we will have two important tools we need for analyzing rotational dynamics. Rotational Kinetic Energy Any moving object has kinetic energy. We know how to calculate this for a body undergoing translational motion, but how about for a rigid body undergoing rotation? This might seem complicated because each point on the rigid body has a different velocity. However, we can make use of angular velocity—which is the same for the entire rigid body—to express the kinetic energy for a rotating object. rotational kinetic energy. Figure 10.17 The rotational kinetic energy of the grindstone is...

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:The kinetic energy KE of an object of mass m moving with velocity v is defined as KE = If a force f(x) acts on the object, moving it along the x-axis from x1 to x2, the Work-Energy Theorem states that the net work done is equal to the change in kinetic energy: mv,2 - mv,?, where vi is the velocity at x1 and v2 is the velocity at x2. Suppose that when The kinetic energy KE of an object of mass m moving with velocity v is defined as KE = If a force f(x) acts on the object, moving it along the x-axis from x1 to x2, the Work-Energy Theorem states that the net work done is equal to the change in kinetic energy: mv,2 - mv,?, where vi is the velocity at x1 and v2 is the velocity at x2. Suppose that when launching a 500-kg roller coaster car an electromagnetic propulsion system exerts a force of 5.7x2 + 1.5x newtons on the car at a distance x meters along the track. Use the Work-Energy Theorem to find the speed of the car when it has traveled 70 meters. (Round your answer to two decimal places.) X m/s Need Help? Talk to a Tutor Read It Previous question Next question

The kinetic energy of a body depends upon mass (m) and velocity (v) of the body. Let K.E. ∝ m x v y ∴ K.E. = km x v y .....(1) where, k = dimensionless constant of proportionality. Taking dimensions on both sides of equation (1), [L 2M 1T -2] = [L 0 M 1 T 0] x [L 1 M 0 T -1] y = [L 0 M x T 0] [L y M 0 T -y] = [L (0+y) M (x + 0) T (0- y)] [L 2 M 1 T -2] = [L y M x T -y]....(2) Equating dimensions of L, M, T on both sides of equation (2), x = 1 and y = 2 Substituting x, y in equation (1), we have K.E. = kmv 2

$$ KE = \dfracmv^2 $$ Where: • KE = kinetic energy • m = mass of a body • v = velocity of a body Kinetic Energy Kinetic Energy is the energy an object has owing to its motion. In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2. We use Joules, kilograms, and meters per second as our defaults, although any appropriate units for mass (grams, ounces, etc.) or velocity (miles per hour, millimeters per second, etc.) could certainly be used as well - the calculation is the same regardless. References/ Further Reading • •