An object was dropped from a height of 20 m from the ground. Use energy terms to find the objects speed when it is 5m from the ground (17m/s) is the answer ​

An object was dropped from a height of 20 m from the ground. Use energy terms to find the objects speed when it is 5m from the ground
(17m/s) is the answer ​

We can use the law of conservation of energy to solve this problem. At the beginning, the object only has potential energy (PE) due to its position above the ground. As it falls, some of this potential energy is converted into kinetic energy (KE) due to its motion. At any point during the fall, the total energy (PE + KE) remains constant.

At the beginning, when it is dropped from a height of 20m, the object has only potential energy given by:

PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the height above ground. Assuming m = 1 kg, we have:

PE = 1 x 9.8 x 20 = 196 J

When the object is at a height of 5m above the ground, it has lost some of its potential energy and gained some kinetic energy. At this point, let's assume it has a speed of v meters per second. Its potential energy is now:

PE = mgh = 1 x 9.8 x 5 = 49 J

Its kinetic energy is:

KE = 0.5mv2

The total energy is:

PE + KE = 196 J (from the beginning) = 49 J + 0.5mv2 (at a height of 5m)

Solving for v, we get:

v = sqrt((196 - 49)/0.5m) = sqrt(294/m) = sqrt(294/1) = 17 m/s

Therefore, the object's speed when it is 5m from the ground is approximately 17m/s.

This is a physics problem that can be solved using the work-energy theorem, which states that the net work done on an object equals the change in its kinetic energy. Kinetic energy is the energy an object has because of its motion and it depends on its mass and speed.

Let's assume that the object has a mass of m and it falls freely under the influence of gravity. The initial height of the object is 20 m and the final height is 5 m. The initial speed of the object is zero and the final speed is v. The work done by gravity on the object is equal to the change in its gravitational potential energy, which is mgh, where g is the acceleration due to gravity and h is the height. The work done by air resistance is assumed to be negligible.

Using the work-energy theorem, we can write:

W_{net} = \Delta K

mgh_{i} - mgh_{f} = \frac{1}{2}mv^{2} - \frac{1}{2}m(0)^{2}

mg(20) - mg(5) = \frac{1}{2}mv^{2}

15mg = \frac{1}{2}mv^{2}

v^{2} = 30g

v = \sqrt{30g}

Therefore, the speed of the object when it is 5 m from the ground is approximately 24.25 m/s (assuming g = 9.8 m/s^{2}).

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