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Consult Conceptual Example 9 in preparation for this problem. The drawing shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 13.0 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 5.20 m above the water? Ignore the effects of air resistance.

8.2$m / s$

Physics 101 Mechanics

Chapter 6

Work and Energy

Work

Kinetic Energy

Simon Fraser University

Hope College

University of Winnipeg

Lectures

03:47

In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

02:08

In physics, work is the transfer of energy by a force acting through a distance. The "work" of a force F on an object that it pushes is defined as the product of the force and the distance through which it moves the object. For example, if a force of 10 newtons (N) acts through a distance of 2 meters (m), then doing 10 joules (J) of work on that object requires exerting a force of 10 N for 2 m. Work is a scalar quantity, meaning that it can be described by a single number-for example, if a force of 3 newtons acts through a distance of 2 meters, then the work done is 6 joules. Work is due to a force acting on a point that is stationary-that is, a point where the force is applied does not move. By Newton's third law, the force of the reaction is equal and opposite to the force of the action, so the point where the force is applied does work on the person applying the force. In the example above, the force of the person pushing the block is 3 N. The force of the block on the person is also 3 N. The difference between the two forces is the work done on the block by the person, which can be calculated as the force of the block times the distance through which it moves, or 3 N × 2 m = 6 J.

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Okay, so let's break down this question into 2 parts, so the first part is from the rest point to the path 1 going to energy conservation of the initial total energy can be equal to the total energy of the path 1. Therefore, we have the initial kinetic energy plus. The initial gravitational potential energy is equal to the king, egypt, 1, plus the capitation potential energy, a path 1. So we have 1 half mv square plus m g. H is equal to 1 half m v, 1 squared plus m g h, 1 point so we know initially, the speed is equal. 0. So therefore, have i equals 0, since the person was at rest and the heights at a pathan should be 0, since the person fall into the water just after the person released the rope, therefore, the height of pethoniso. Now we have m g, h is equal to 1 half m v 1 square, so, as you can tell mass m can cancel out and it will give us g h. I is equal. 1 have 1 square. So now we have h. I is equal to 1 square over 2 g p, so we know 1 here, which is the speed of the path 1 is given as 13 meters per second and g. Is the gravitational constant? I'M sorry, no gravitational constant. The acceleration of the gravity, which is 9.8 meter per second square, so now we can plug in these various to the equation. Eventually, we'll have the initial height, which is the high at starting point, is equal to 8.62 meter. So now, let's take a look at the second part, so the second part is, from the rest point to path 2, so we can still apply the energy conservation al to have the initial total energy is equal to the total energy. It has 2, and this will give us 1 half m v square plus m g. H is equal to 1 half m v 2 square plus m g h 2, as you can tell mass m, can we can sell out, and this will give us 1 half v i square plus g h. I is equal to 1 half v 2 square plus g h 2 point so i lose our ransom here have v. 2 is equal to square root. I squared plus 2 g times h. I minus h 2 point so since we know the initial speed is equal to 0, so v 2 is equal to square root, 2 g times h, minus h, 2 point. So we know the acceleration of gravity is 9.8 meter per second square. In the initial height. Is 8.62 and the height at pat 2 is given as 5.2 meter so now his plug in his values bank to the question to determine the speed at pas 2, which is equal to square root, 2 times 9.8 meter per second square and 10 times 8.62 meter. Minus 5.2 meter- and this is equal to 8.2 meter per second- and this is the answer for this quest.

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