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2 years ago

5

You pick up a 3.4 kg can of paint from the ground and lift it to a height of 1.8 m. how much work do you do on the can of paint?

you hold the can stationary for half a minute, waiting for a friend on a ladder to take it. how much work do you do during this time? you friend decides against the paint, so you lower it back to the ground. how much work do you do on the can as you lower it?

1 answer:

zepelin [54]2 years ago

5 0

1) +60.0 J

The work done when lifting an object is equal to the change in gravitational potential energy of the object:

$W=\Delta U= m g \Delta h$

where m is the mass of the object, g is the gravitational acceleration, $\Delta h$ is the variation of height of the object.

In this problem, m = 3.4 kg, g = 9.8 m/s^2 and $\Delta h = +1.8 m$ , so the work done is:

$W=(3.4 kg)(9.8 m/s^2)(1.8 m)=+60.0 J$

And the work done is positive, since the object has gained potential energy.

2) 0 J

In this case, there is no variation of the heigth of the object:

$\Delta h =0$

Therefore, the variation of potential energy is zero:

$\Delta U = 0$

And so, the work done is zero as well:

$W=0$

3) -60.0 J

In this case, the variation of height of the object is negative, since the object has been lowered to the ground:

$\Delta h = -1.8 m$

So, the object has lost potential energy:

$W=(3.4 kg)(9.8 m/s^2)(-1.8 m)=-60.0 J$

And so the work done is negative:

$W=-60.0 J$

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Answer:

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Explanation:

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So we can use a model to describe the density as function of the radius

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So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

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Simplyfind this last expression we have:

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$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

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