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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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3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

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A silver sphere with radius 1.3611 cm at 23.0°C must slip through a brass ring that has an internal radius of 1.3590 cm at the s
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Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

<u>The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.</u>

<u>Thermal expansion for brass = 19×10⁻⁶ °C⁻¹</u>

Thus,

\alpha=\frac {\Delta r}{r\times \Delta T}

Also,

\Delta T=T_{final}-T_{Initial}

So,

19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}

Solving for final temperature as:

(T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}

<u>Final temperature = 106.165 °C</u>

5 0
3 years ago
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