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3 years ago

15

A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How

much work was done by the forklift to lift the boxes?

1 answer:

Nata [24]3 years ago

8 0

Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

$W=F.d.Cos \alpha$

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

$F=1000N.2m.Cos(0)=2000N.m=2000 J$

Have a nice day!

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Answer:

<em>The Volume is 5.018 cubic units</em>

Explanation:

<u>Volume Of A Solid Of Revolution</u>

Let f(x) be a continuous function defined in an interval [a,b], if we take the area enclosed by f(x) between x=a, x=b and revolve it around the x-axis, we get a solid whose volume can be computed as

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And the boundaries defined as x=1, y=0 and revolved around the x-axis. The left endpoint of the integral is easily identified as x=0, because it defines the beginning of the region to revolve. So we need to compute

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We pick $u=x^2,\ dv=e^{2x}dx$

Then $du=2xdx,\ v=\frac{e^{2x}}{2}$

Applying by parts:

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$\displaystyle I_1=x\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}dx$

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$\displaystyle I=\frac{x^2e^{2x}}{2}-\left(\frac{xe^{2x}}{2}-\frac{1}{2}\frac{e^{2x}}{2}\right)$

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$\displaystyle I=\dfrac{\left(2x^2-2x+1\right)\mathrm{e}^{2x}}{4}$

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$V=\pi \left[\dfrac{\left(2(1)^2-2(1)+1\right)\mathrm{e}^{2(1)}-\left(2(0)^2-2(0)+1\right)\mathrm{e}^{2(0)}}{4}\right]$

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