A shopper in a supermarket pushes a cart with a force of 35 N directed

Yolanda Jorge 2021-11-22 Answered
A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of \(\displaystyle{25}^{{\circ}}\) below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. Find the work done by the shopper as she moves down a \(\displaystyle{50.0}-{m}\) length aisle.

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Expert Answer

Elizabeth Witte
Answered 2021-11-23 Author has 6725 answers
Givens:
\(\displaystyle{F}_{{{a}{p}{p}}}={35}{N}\)
\(\displaystyle\theta={25}^{{\circ}}\) Down the horizontal.
\(\displaystyle\triangle{x}={50}{m}\)
We know that the work law is: \(\displaystyle{W}={F}{\cos{\theta}}{d}={F}_{{{p}{a}{r}{a}{l}\le{l}}}{d}\)
In this case \(\displaystyle{F}_{{{p}{a}{r}{a}{l}\le{l}}}={F}_{{{a}{p}{p}}}{\cos{{25}}}°\)
\(\displaystyle{W}_{{{m}{a}{n}}}={F}_{{{a}{p}{p}}}{\cos{{25}}}°\triangle{x}={35}{x}\times{\cos{{25}}}°\times{50}\)
\(\displaystyle{W}_{{{m}{a}{n}}}={1.58}{k}{J}\)
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Jennifer Hill
Answered 2021-11-24 Author has 6006 answers
The expression for the work done by the shopper in moving the cart down the aisle is given by,
\(\displaystyle{W}={F}\cdot{d}\)
PSKW=Fd\cos\theta
\(\displaystyle{W}={35}{N}\times{50}{m}\times{\cos{{25}}}°\)
\(\displaystyle{W}={1586.04}{J}\)
Therefore, the work done by the shopper in moving the cart down the aisle is 1586.04 J.
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