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AlladinOne [14]

3 years ago

13

As a football player moves in a straight line [displacement (3.00 mm)i^i^ - (6.50 mm)j^j^], an opponent exerts a constant force

(126 NN)i^i^ (168 NN)j^j^ on him. How much work does the opponent do on the football player

2 answers:

LuckyWell [14K]3 years ago

6 0

Answer:

0.714 Joule

Explanation:

Work done is defined as the dot product of force vector and the displacement vector.

$W=\overrightarrow{F}.\overrightarrow{d}$

Here, $\overrightarrow{d} = (3\widehat{i}-6.5\widehat{j})mm=(0.003\widehat{i}-0.0065\widehat{j})m$

$\overrightarrow{F}=(126\widehat{i}+168\widehat{j})N$

$W=(126\widehat{i}+168\widehat{j}).(0.003\widehat{i}-0.0065\widehat{j})$

W = 0.378 - 1.092

W = - 0.714 Joule

Thus, the work is 0.714 Joule.

Brilliant_brown [7]3 years ago

4 0

Answer:

Work done, $W=(0.378i-1.092j)\ J$

Explanation:

Displacement,

$d=(3i-6.5j)\ mm\\\\d=(0.003i-0.0065j)\ m$

Force, $F=(126i+168j)\ N$

Work done by the opponent do on the football player is given by :

$W=F{\cdot} d\\\\W=(126i+168j){\cdot} (0.003i-0.0065j)\ m\\\\W=(0.378i-1.092j)\ J$

So, the work done by the opponent do on the football player is $(0.378i-1.092j)\ J$ .

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I hope it helps you!

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