Write about the similarities and differences between kinetic and potential energy. Include specific, real world example in your
explanation.
1 answer:
Answer:
Energy is transformed from potential to kinetic and vice versa
Explanation:
The energy is transformed from mechanical to kinetic energy when the object changes its position with respect to a reference point, where it loses height but increases its speed. When the object is at maximum height with respect to a reference point, it will have its maximum potential energy value. When the object passes through the reference point it will have potential energy equal to zero, but this energy will become kinetic energy.
The most characteristic and real example is that of a pendulum at one end, as can be seen in the attached image.
When the pendulum is located at the top end, as shown in Figure 1, at that point the maximum potential energy will be held. Then the pendulum is released and when it passes through the reference point and its height is zero, with respect to that point, all potential energy will have become kinetic energy in the same way at this point the maximum speed of the pendulum will be set.
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a) Solids keep shape, liquids take shape of containers but don't spill, gases take container's shape and spill out
b) if you heat gas, speed of its molecules will increase and they'll push the container's walls stronger, so the pressure will increase when the container heated
c) Heat flows from warmer to colder bodies
d) For monatomic gases it's U=1.5nRT only, molecular gas has bonds between atoms so total internal energy increases
e) Of gases
f) m/s
g) U=5/2*nRT=37830.85 J
Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
- If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:
- Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
- Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
- τ = F*r (2)
- For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
- Replacing (2) and (3) in (1), we can solve for α, as follows:
- Since the angular acceleration is constant, we can use the following kinematic equation:
- Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:
- Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:
- Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:
- where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
- Replacing this value and (7) in (8), we get:
b)
- There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:
- where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
- Replacing this value and (4), in (9), we get:
- Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
- The radial acceleration is just the centripetal acceleration, that can be expressed as follows:
- Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
- Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
- Replacing this value and (7) in (11) we get:
- The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
- Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb