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steposvetlana [31]
3 years ago
6

2. A jack exerts a vertical force of 4.5 X 103

Physics
1 answer:
skad [1K]3 years ago
5 0

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

\\  \\

Given :-

\star  \sf  \small force = 4.5 \times  {10}^{3}  \: newton

\star  \sf  \small distance = 0.25 \: meter

\\  \\

To find:-

\sf \star \: work = \: ?

\\  \\

Solution:-

we know :-

\bf \dag \boxed{ \rm work = force \times distance}

\\  \\

So:-

\dashrightarrow \sf work = force \times distance

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{0 \cancel.5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \cancel \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work =  \dfrac{4\cancel.5}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{3 - 1} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{1}  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times  \cancel{10}}{ \cancel2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times 5}{ 1} \\

\\  \\

\dashrightarrow \sf work =225 \times 10

\\  \\

\dashrightarrow \bf work =\red{2250\: joule}

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