Physics, 24.12.2019 06:31 mimithurmond03

A5 kg object near earth's surface is released from rest such that it falls a distance of 10 m. after the object falls 10 m, it has a speed of 12 m/s. which of the following correctly identifies whether the object-earth system is open or closed and describes the net external force?

a) the system is closed, and the net external force is zero.
b) the system is open, and the net external force is zero.
c) the system is closed, and the net external force is nonzero.
d)the system is open, and the net external force is nonzero.

Answers: 2

Show answers

Answers

To solve the problem it is necessary to apply the concepts related to energy conservation. We have a mass of 5Kg that at a distance of 10m has gained 12m / s, with energy losses due to friction.

KE + W_f = PE

$\frac{1}{2}mv^2 +W_f = mgh$

$W_f = \frac{1}{2}mv^2 -mgh$

$W_f = \frac{1}{2} 5*12^2-5*9.8*10$

W_f = 130J

Since there is a loss of energy due to friction, it means that an external force resulting from friction is interacting on the system, which also makes the system open.

The correct answer is D.

Answer from: jaimechristopher2002

There will be some energy loss due to friction countering the non-zero value of external force, which makes the system open. Hence, option (D) is correct.

Given data:

The mass of object is, m = 5 kg.

The falling distance is, h = 10 m.

Speed of object after falling the distance of 10 m is, v = 12 m/s.

This problem can be solved with an approach of concepts related to the energy conservation. We have a mass of 5Kg that at a distance of 10m has gained 12m / s, with energy losses due to friction.

Then as per the conservation of Energy,

PE = KE + W

Here,

PE is the potential energy.

KE is the kinetic energy.

W is the energy losses due to friction.

Solving as,

$PE = KE + W\\\\W=PE - KE\\\\W =mgh-\dfrac{1}{2}mv^{2}\\\\W =(5 \times 9.8 \times 10)-\dfrac{1}{2} \times 5 \times 12^{2}\\\\\\W = 130 \;\rm J$

So, there will be energy loss of 130 J due to friction, which ultimately means that an external force resulting from friction is interacting on the system. And if there is an interaction of system with surrounding parameters (here it is external force), then the system is open.

Thus, we can conclude that there will be some energy loss due to friction countering the non-zero value of external force, which makes the system open. Hence, option (D) is correct.

Learn more about the conservation of energy here:

link

Answer from: Quest

they are playing too much beyblade in the office

explanation:

Answer from: Quest

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

Look at the pendulum diagram between which points does the pendulum’s potential energy increase most

Another question on Physics

Physics, 21.06.2019 17:20

When an object moves in uniform circular motion, the direction of its acceleration is a) in the same direction as its velocity vector. b) in the opposite direction of its velocity vector c) is directed toward the center of its circular path. d) is directed away from the center of its circular path. e) depends on the speed of the object.

Answers: 1

Answer

Physics, 22.06.2019 05:20

Alfred pushes on a heavy box, but cannot move it. the box has a lot of inertia motion friction gravity

Answers: 1

Answer

Physics, 22.06.2019 07:00

The seven sisters, seven stars located more than 400 light-years away in the taurus constellation can be seen here with venus and another constellation, orion. the seven stars stay grouped together but seem to move, like the planet venus. why do stars seem to move like planets in the night sky?

Answers: 3

Answer

Physics, 22.06.2019 11:00

How do we measure or think of speed in the united states?

Answers: 1

Answer

You know the right answer?

A5 kg object near earth's surface is released from rest such that it falls a distance of 10 m. after...

Questions