5) Why does the first hill/drop on a roller coaster have to be the highest? (think about
potential and kinetic energy)
6) How much work does it take to lift an object weighing 200 N to a distance of 20 meters?
7) 300 J of work are needed to push an object with a force of 75 N. What is the distance
traveled?
8) Calculate the potential energy of a bird that has a mass of 5 kg and it is 50 meters high?

Respuesta :

5) Due to the conservation of energy / due to the presence of friction

6) The work done is 4000 J

7) The distance travelled by the object is 4 m

8) The potential energy of the bird is 2450 J

Explanation:

5)

For a roller coaster in motion along the track, in absence of friction, the mechanical energy remains constant:

[tex]E=U+K = const.[/tex]

where

U is the potential energy

K is the kinetic energy

At the beginning, the roller coaster is at rest, therefore its kinetic energy is zero and its mechanical energy is just equal to the potential energy:

[tex]E=U=mgh[/tex]

where m is the mass of the roller coaster, g the acceleration of gravity, and h the height of the roller coaster above the ground at the first hill.

Since this amount of energy remains constant, the roller coaster cannot go to a higher hill: in fact, in that case it would reach a height [tex]h'>h[/tex], but this is not possible, because it would mean that its mechanical energy would be [tex]mgh' > E[/tex], larger than its mechanical energy, and since energy cannot be created (law of conservation of energy), the height of the next hills cannot be higher than the first one. Moreover, if we consider friction, then friction does work on the roller coaster in motion, and as a result, part of its mechanical energy is wasted (converted into thermal energy): therefore, the height of the following hills must be lower than the height of the first hill.

6)

The work done when lifting an object is equal to the gravitational potential energy gained by the object:

[tex]W=Fh[/tex]

where

F is the weight of the object

h is the change in height of the object

For the object in this problem, we have

F = 200 N (weight)

h = 20 m (change in height)

Substituting, we find the work done:

[tex]W=(200)(20)=4000 J[/tex]

7)

The work done when pushing an object in a direction parallel to the motion of the object is given by:

[tex]W=Fd[/tex]

where

F is the magnitude of the force

d is the distance through which the object has travelled through

In this problem, we have:

W = 300 J (work done)

F = 75 N (magnitude of the force)

Solving the equation for d, we find the distance travelled:

[tex]d=\frac{W}{F}=\frac{300}{75}=4 m[/tex]

8)

The potential energy of an object is the energy possessed by an object due to its position in a gravitational field. Near the Earth's surface, it is given by

[tex]PE=mgh[/tex]

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 bird in this problem, we have:

m = 5 kg (mass)

[tex]g=9.8 m/s^2[/tex]

h = 50 m (height)

Substituting, we find:

[tex]PE=(5)(9.8)(50)=2450 J[/tex]

Learn more about potential energy:

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