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3 years ago

A ball is dropped from the top of a building.

Physics

2 answers:

son4ous [18]3 years ago

8 0

Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point

olga nikolaevna [1]3 years ago

5 0

Answer:

Right before it hits the ground.

Explanation:

It has the most kinetic energy at the very end of its descent when it is moving the fastest.

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A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

A year 11 pupil with a mass of 55kg swinging back on their chair and falling off it at a speed of 0.6m/s. What is his kinetic en

posledela

Answer:

Uk = 9.9 J

Explanation:

To calculate the kinetic energie (Uk), you can make use of this formula:

Uk = 0.5 * m * v²

given m = 55 kg and v = 0.6 m/s

Substituting in the formula gives:

Uk = 0.5 * 55 * (0.6)²

Uk = 0.5 * 55 * 0.36

Uk = 9.9 J

Extra:

Now let's examine the formula in relation to the SI units. If you understand the following, it will give you great insight in how smart Phisics is inter twained by looking at formulas and their standard units. It will save you time in future to convert formulas, if you use the right standard units.

The formula for kinetic energie is:

Uk = 0.5 * m * v²

Standard SI unit for mass m is kg.

Standard SI unit for speed v is m/s.

So v * v = v² and therefore v² must have the standard SI unit of m²/s².

From the formula, you see that the unit of Uk must be kg*m²/s² and since Uk is normally given in J, these both forms must be the same !

The main unit for Uk is the Joule. Now please see the picture, which shows the relation between the J and other SI units. Please understand that you can construct your 'own' formulas based these units. Now here is the time saver:

Because almost always the right units are given in a question, or because sometimes you can look up a constant in a table with an exotic and seemingly complicated unit, but that says a lot about the formula which must have been some how involved!

By this, I hope you now understand the implication of using the right standard SI units and how that can help you figure out what formula is needed.

3 0

2 years ago

Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ

elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level = KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0

3 years ago

a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time

Evgen [1.6K]

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

$S=ut + \frac{1}{2}at^2$