Question

How much work does it take to slide a crate m along a loading dock by pulling on it with a -n force at an angle of from the horizontal?

Answer 1

The work needed to slide the crate at a distance of 22 m is

$W=4.38 \times 10^3 J$

What is meant by work done?

The work done by a constant force on an object as the object moves along a displacement vector $\vec{d}$ is given by the formula involving the dot product of the two vectors:

$W=\vec{F} \cdot \vec{d}=F d \cos (\theta)$

Where $\vec{F}$, F is the force vector and its magnitude respectively, $\vec{d}$, d is the displacement vector and its magnitude respectively and $\theta$ is the angle between the displacement vector and the force vector.

Given:

Force acting on the crate, F = 230 N

Distance traveled by the crate,

$d=22 \mathrm{~m}$

The angle between the force and the displacement,

$\theta=30^{\circ}$

Then the work done by the force on the crate as it travels 22 m under the action of the force is

$W &=F d \cos \theta$

$&=230 \times 22 \cos 30^{\circ}$

simplifying the above equation, we get

$&=230 \times 22 \times \frac{\sqrt{3}}{2}$

$&=4382.089 \mathrm{~J}$

$&=4.38 \times 10^3 \mathrm{~J}$

Therefore the work needed to slide the crate at a distance of 22 m is

$W=4.38 \times 10^3 J$

The complete question is:

How much work does it take to slide a crate 22m along a loading dock by pulling on it with a $230-\mathrm{N}$ force at an angle of $30^{\circ}$ from the horizontal?

To learn more about work done refer to:

brainly.com/question/25573309

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