Paloma Owens

2023-03-30

A consumer in a grocery store pushes a cart with a force of 35 N directed at an angle of ${25}^{\circ }$ below the horizontal. The force is just enough to overcome various frictional forces, so the cart moves at a steady pace. Find the work done by the shopper as she moves down a $50.0-m$ length aisle.

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To find the work done by the shopper as she moves down the aisle, we can use the formula for work:
$W=F·d·\mathrm{cos}\left(\theta \right)$
Where:
$W$ is the work done,
$F$ is the applied force,
$d$ is the distance, and
$\theta$ is the angle between the force and the direction of motion.
In this case, the applied force is 35 N and the distance is 50.0 m. The angle between the force and the direction of motion is 25 degrees below the horizontal.
Substituting these values into the formula, we get:
$W=35\phantom{\rule{0.167em}{0ex}}\text{N}·50.0\phantom{\rule{0.167em}{0ex}}\text{m}·\mathrm{cos}\left({25}^{\circ }\right)$
To calculate the cosine of 25 degrees, we need to convert the angle to radians. We know that $\pi$ radians is equal to 180 degrees, so we can convert 25 degrees to radians by multiplying it by $\frac{\pi }{180}$:
$W=35\phantom{\rule{0.167em}{0ex}}\text{N}·50.0\phantom{\rule{0.167em}{0ex}}\text{m}·\mathrm{cos}\left(\frac{25\pi }{180}\right)$
Now, we can calculate the cosine of the angle:
$W=35\phantom{\rule{0.167em}{0ex}}\text{N}·50.0\phantom{\rule{0.167em}{0ex}}\text{m}·\mathrm{cos}\left(\frac{25\pi }{180}\right)\approx 35\phantom{\rule{0.167em}{0ex}}\text{N}·50.0\phantom{\rule{0.167em}{0ex}}\text{m}·0.90631$
Calculating the product of the force, distance, and cosine value:
$W\approx 1581.55\phantom{\rule{0.167em}{0ex}}\text{N}·\text{m}\approx 1581.55\phantom{\rule{0.167em}{0ex}}\text{J}$
Finally, we can convert the work from joules to kilojoules by dividing by 1000:
$W\approx \frac{1581.55\phantom{\rule{0.167em}{0ex}}\text{J}}{1000}\approx 1.58\phantom{\rule{0.167em}{0ex}}\text{kJ}$
Therefore, the work done by the shopper as she moves down the 50.0 m length aisle is approximately $1.58\phantom{\rule{0.167em}{0ex}}\text{kJ}$, which can be represented as ${W}_{\text{man}}=1.58\phantom{\rule{0.167em}{0ex}}\text{kJ}$.

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