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aleksandr82 [10.1K]
3 years ago
14

In which of these situations, is mechanical energy being conserved? (Neglect, air resistance, friction, and breaking) Check all

that apply.
Child on a swing


Pendulum


Bow and Arrow


Roller Coaster

Question 2
As an object falls to the ground, its potential energy is being converted to kinetic energy.


True
False
Question 3
The energy that an object has stored due to its position or shape is called:

kinetic energy


potential energy


chemical energy


electrical energy

Question 4
Calculate the potential energy stored in a 10 kg block, 5 meters above the ground.


490 J


50 J


490 N


50 N

Question 5
An elevator weighs 1500 Newtons. Calculate how much potential energy it has when it is lived 500 meters in the air.


75,000 J


750 J


7,500,000 J


750,000 J
Physics
1 answer:
lana66690 [7]3 years ago
5 0

1) Mechanical energy is conserved in all the situations listed

2) True

3) The energy that an object has stored due to its position or shape is called potential energy

4) The potential energy of the block is 490 J

5) The potential energy of the elevator is 750,000 J

Explanation:

1)

The mechanical energy of an object is the sum of its kinetic energy (KE) and its potential energy (PE):

E=KE+PE

Where

KE is the energy due to the motion of the object

PE is the energy due to the position of the object (it can be either gravitational potential energy or elastic potential energy)

In absence of non-conservative forces, such as friction or air resistance, the mechanical energy is always conserved. Therefore, the mechanical energy is conserved in all the situations listed here:

Child on a swing  --> there is a continuous conversion between gravitational potential energy and kinetic energy

Pendulum  --> there is a continuous conversion between gravitational potential energy and kinetic energy

Bow and Arrow  --> there is a conversion between elastic potential energy of the bow and kinetic energy of the  arrow

Roller Coaster --> there is a continuous conversion between gravitational potential energy and kinetic energy

2)

The potential energy of an object is given by

PE=mgh

where

m is its mass

g is the acceleration due to gravity

h is the height of the object relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where

v is the speed of the object

As an object falls to the ground, its height h decreases, therefore the potential energy PE decreases as well. However, the speed of the object, v, increases during the fall, and therefore the kinetic energy KE increases. This means that potential energy is converted into kinetic energy.

3)

Potential energy is the energy possessed by an object due to its position. It can be of two types:

  • Gravitational potential energy: it is the potential energy due to the position of an object in a gravitational field. It is calculated as mgh, as shown in part 2)
  • Elastic potential energy: it is the potential energy stored in an elastic object when it is stretched or compressed. It is calculated as \frac{1}{2}kx^2, where k is the spring constant of the elastic object and x is the stretching/compression of the object relative to its equilibrium position.

4)

The potential energy stored in an object held above the ground is given by

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object relative to the ground

For the object in this problem, we have

m = 10 kg

g=9.8 m/s^2

h = 5 m

Substituting, we find

PE=(10)(9.8)(5)=490 J

5)

As before, the potential energy of the elevator is given by

PE=mgh

where m is its mass and h is its height above the ground.

Here we don't have the mass of the elevator. However, we know its weight:

W=1500 N

But we also know that the weight of an object is equal to the product between its mass and the acceleration of gravity:

W=mg

So we can rewrite the potential energy as

PE=Wh

and the height of the elevator is

h = 500 m

Therefore, its potential energy is

PE=(1500)(500)=750,000 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

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