Potential Problems in Energy Practice - GeeksforGeeks (2023)

In everyday language, the word potential is often used for things or people that are promising in themselves. "Potential" shows the possibility of action. It gives an idea of ​​the stored energy that can be converted. That's the idea behind potential energy. This concept is integral to mechanics and allows us to theoretically measure the energy stored in an object. Potential energy can come through any force. For example, a stretched or compressed spring has potential energy. An object that is at a certain height has potential energy because of the height.

potential energy

Potential energy is the energy an object possesses due to its position or configuration. This energy should be stored in the object. Normally, potential energy is released from an object through motion. For example, when a stretched spring is released, it begins to move to its natural position and begins to increase in speed. Because of this speed, it gains kinetic energy. To better understand this, let's consider a dumpling "m".

This ball, which initially lies on the ground, rises to a height of "h". The external force acts on it in the form of gravity. We know that the work done by a force F at a displacement "s" on the object is given by

W = Fs

The force in this case is gravity and the displacement of the ground up to the height "h".

F = mg, s = h

Then the work done by gravity on the object is

W = -mg

Potential energy is defined as the negative of this work. Denotes the potential energy V(h).

V(h) = mgh

In this case, when the ball falls, the potential energy decreases and the speed increases. This means that the potential energy of the object is converted into kinetic energy. Suppose the speed of the ball just before it hits the ground is "v".

potential energy in a spring

When the spring is held normally, it has 0 energy and is in balance. When it is stretched or compressed and there is a certain displacement, say x, it has stored some potential energy, which is given as

P.E.= 1/2 (Kx²)

Where K = spring constant

x = displacement due to compression or expansion.

Let's look at some problems based on these concepts.

examples of problems

Task 1: A mass of 2 kg is lifted from the ground to a height of 10 m. Find the potential energy of the object.

Answer:

The potential energy of a mass "m" at height "h" is given by

P = mgh

Data: m = 2 kg and g = 10 m/s²e = 10m.

Objective: Find the potential energy.

Inserting the values ​​into the formula.

P = mgh

⇒ P = (2)(10)(10)

⇒ P = 200J

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So the potential energy of the object is 200 J.

Task 2: A mass of 5 kg is lifted from the ground to a height of 100 m. Find the potential energy of the object.

Answer:

The potential energy of a mass "m" at height "h" is given by

P = mgh

Data: m = 5 kg and g = 10 m/s²e = 100m.

Objective: Find the potential energy.

Inserting the values ​​into the formula.

P = mgh

⇒ P = (5)(10)(100)

⇒P = 5000J

So the potential energy of the object is 5000 J.

Task 3: A mass weighing 5 kg is lifted from the ground 5 m high onto the wedge. The wedge forms an angle of 30° with the ground. Find the potential energy of the block.

Answer:

The potential energy of a mass "m" at height "h" is given by

P = mgh

This wedge is in the shape of a right triangle.

Figure

We say,his the vertical height reached by the box, be the inclined lengthUE

large = 5m

Data: m = 5 kg and g = 10 m/s²h = 2,5 m.

Objective: Find the potential energy.

Inserting the values ​​into the formula.

P = mgh

⇒ P = (5)(10)(2,5)

⇒P = 125J

So the potential energy of the object is 125 J.

Task 4: A 10 kg mass is lifted from the ground 10 m high onto the wedge. The wedge forms an angle of 30° with it even. Find the potential energy of the block.

Answer:

The potential energy of a mass "m" at height "h" is given by

P = mgh

This wedge is in the shape of a right triangle.

Figure

We say,his the vertical height reached by the box, be the inclined lengthUE

large = 10m

Data: m = 5 kg and g = 10 m/s²is = 5m.

Objective: Find the potential energy.

Inserting the values ​​into the formula.

P = mgh

⇒ P = (5)(10)(5)

⇒P = 250J

So the potential energy of the object is 250 J.

Question 5: Find the kinetic energy of the ball just before it hits the ground. Suppose the sphere is initially 10 m high and has a mass of 2 kg.

Answer:

At a height of 10 m, the sphere initially has potential energy. As it falls, it begins to descend toward the ground and its height begins to decrease. As altitude decreases, speed increases and it gains kinetic energy.

Potential energy at t = 0

The potential energy is given by,

P = mgh

m = 2 kg, h = 10 m und g = 10 m/s²

P = mgh

⇒ P = (2)(10)(10)

⇒ P = 200J

When the ball is about to hit the ground, its potential energy becomes zero and all energy is converted to kinetic energy.

Also KE = 200J

Question 6: Find the speed of the ball just before it hits the ground. Assume that the sphere is initially 100 m high and has a mass of 4 kg.

Answer:

At a height of 10 m, the sphere initially has potential energy. As it falls, it begins to descend toward the ground and its height begins to decrease. As altitude decreases, speed increases and it gains kinetic energy.

Potential energy at t = 0

The potential energy is given by,

P = mgh

m = 4 kg, h = 100 m und g = 10 m/s²

P = mgh

⇒ P = (4)(100)(10)

⇒P = 4000J

When the ball is about to hit the ground, its potential energy becomes zero and all energy is converted to kinetic energy.

Also KE = 4000J

The formula for K.E is:

KE =

m = 4 kg and v = ?. Plug the values ​​into the formula

KE =

⇒ 4000 =

⇒2000 =v²

⇒ v = 10√20 m/s

⇒ v = 20√5 m/s

Question 7: All of a ball's potential energy is converted to its kinetic energy when it hits the ground from a certain height. The height at which the ball was originally placed was 10 m. The mass of the ball is 1 kg. Find the kinetic energy gain.

Solution:

Since all the potential energy present in the ball is transferred to its kinetic energy,

Potential energy of the ball = final gain in kinetic energy

P = mgh

m= 1kg, h= 10m, g= 9,8m/sek²

P = 1 × 10 × 9,8

P = 98 Joule

Therefore, the final gain in kinetic energy is 98 joules.

KE = 98. Jul

Question 8: Explain the existence of potential energy,

a. Because of his position

B. by the state in which the object is found.

Answer:

Potential energy can actually exist in two different cases,

a. Because of his position

Suppose an object is standing stable on the ground and is now using some energy to bring it to a certain height. The object at a given altitude will have stored energy in the form of potential energy. It is given as

PE = mgh

B. by the state in which the object is found.

When a spring is held in a normal state it has 0 energy, but when the same spring is compressed or stretched it gains potential energy expressed as:

BODY EDUCATION = 1/2 (Kx²)

Where x = displacement, K = spring constant.

Question 9: A spring is stretched to 9 cm, the elastic constant of the spring is 2 N/m. Find the value of the potential energy stored in the spring?

Solution:

The potential energy stored in a spring is given as

P.E= 1/2 (Kx²)

K= 2 N/m, x= 9cm= 0,09m

PE= 1/2 (2× (0,09)²)

PE= 8,1 × 10^-3Joule.

Question 10: Two objects are held at different heights, the first object is 5 meters away and the second object is 15 meters away. If the first object is five times heavier than the second, which object has more potential energy?

Solution:

Object 1:

Height = 5m, mass = 5m

PHYSICAL EDUCATION₁= 5 m × 5 × 9,8

PHYSICAL EDUCATION₁= 245 million joules.

Object 2:

Height = 15 m, mass = m

PHYSICAL EDUCATION₂= 15 × m × 9,8

PHYSICAL EDUCATION₂= 147 million joules.

Therefore, even if the first object is held at a lower height, the first object has more potential energy due to its weight.

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