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kykrilka [37]
3 years ago
9

You are in a room in a basement with a smooth concrete floor (friction force equals 40 N) and a nice rug (friction force equals

55 N) that is 3 m by 4 m. However, you have to push a very heavy box from one corner of the rug to the opposite corner of the rug. Will you do more work against friction going around the floor or across the rug, and how much extra?
Physics
1 answer:
Musya8 [376]3 years ago
6 0

Answer:

I will do more work against friction going around the floor.

ΔW = 5 J

Explanation:

Given

Ff₁ = 40 N

Ff₂ = 55 N

a = 3 m

b = 4 m

We have to get the distance <em>d₁</em> as follows

d₁ = a + b = 3 m + 4 m = 7 m

And  <em>d₂:</em>

d₂ = √(a² + b²) = √((3 m)² + (4 m)²)

⇒   d₂ = 5 m

then we use the equations

W₁ = Ff₁*d₁ = (40 N)(7 m) = 280 J   (work against friction going around the floor)

W₂ = Ff₂*d₂ = (55 N)(5 m) = 275 J  (work against friction going across the rug)

since  W₂ < W₁  I will do more work against friction going around the floor.

Now, we apply the formula

ΔW = W₁ - W₂ = 280 J - 275 J = 5 J

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Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

F_{g0}=G \frac{mM_{E0}}{r^{2}}

So if the mass of the earth is increased by a factor of 2, this means that:

M_{Ef}=2M_{E0}

so:

F_{gf}=G \frac{2mM_{E0}}{r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=2

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

M_{Ef}=3M_{E0}

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F_{gf}=G \frac{3mM_{E0}}{r^{2}}

Therefore:

\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=3

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

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so:

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\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}

When simplifying we end up with:

\frac{F_{gf}}{F_{g0}}=\frac{1}{4}

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

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