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Tanya [424]
2 years ago
6

A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13

0N . The coefficient of friction between box and floor is 0.300 . Find(d) the work done by the gravitational force..
Physics
1 answer:
Irina18 [472]2 years ago
3 0

The work done by the gravitational force = 0

Given the mass of the box = 40 kg

The box is initially at rest.

Distance moved by the applied force = 5m

Force applied = 130 N

Co-efficient of friction between the box and floor = 0.3

The box is moved only in the horizontal direction by the applied force.

Gravitational force is applied in a direction perpendicular to the applied force. hence it doesn't do any work on the box.

Therefore, the work done by the gravitational force is zero.

Learn more about the gravitational force at brainly.com/question/862529

#SPJ4

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<h3>What is gravitational potential energy?</h3>

Gravitational energy or gravitational potential energy is the potential energy a huge item has corresponding to one more monstrous article because of gravity. It is the potential energy related with the gravitational field, which is delivered (changed over into active energy) when the articles fall towards one another. Gravitational potential energy increments when two articles are brought further apart. It is the potential energy related with the gravitational field, which is delivered (changed over into dynamic energy) when the items fall towards one another. Gravitational potential energy increments when two items are brought further separated.

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Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some,
astraxan [27]

I have to say, i love this kind of problems.

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the angular frequency

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Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

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The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

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We are almost there. If this were any other problem, we would write:

a = A r \omega^2

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Explanation:

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Explanation:

HOPE THIS HELPS!!!!!!❤️

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