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nirvana33 [79]
3 years ago
8

Read through the scenarios below and calculate the predicted change in kinetic energy of the object compared to a 100 kg ball tr

aveling at 5 m/s. A 1,000 kg ball traveling at 5 m/s would have kinetic energy. A 10 kg ball traveling at 5 m/s would have kinetic energy. A 100 kg person falling at 5 m/s would have kinetic energy.
Physics
2 answers:
sashaice [31]3 years ago
4 0

Okay, first off, the formula for Kinetic Energy is:

<em>KE = 1/2(m)(v)^2</em>

<em>m = mass</em>

<em>v = velcoity (m/s)</em>


Using this formula, we can then calculate the kinetic energy in each scenario:

1) KE = 1/2(100)(5)^2 = 1,250 J

2) KE = 1/2(1000)(5)^2 = 12,500 J

3) KE = 1/2(10)(5)^2 = 125 J

4) KE = 1/2(100)(5)^2 = 1,250 J

gregori [183]3 years ago
3 0

Answer:

A 1,000 kg ball traveling at 5 m/s would have <u>10 times more</u> kinetic energy.

A 10 kg ball traveling at 5 m/s would have <u>10 times less</u> kinetic energy.

A 100 kg person falling at 5 m/s would <u>have the same</u> kinetic energy.

Explanation:

Edg 2020

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(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

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F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

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(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

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As before, the work done by any force on the crate is

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We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

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The total work done on the crate is the sum of the work done by the four forces acting on it, so:

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And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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