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?
[tex] \\ \\ [/tex]
Given :-
[tex] \star \sf \small force = 4.5 \times {10}^{3} \: newton[/tex]
[tex] \star \sf \small distance = 0.25 \: meter[/tex]
[tex] \\ \\ [/tex]
To find:-
[tex] \sf \star \: work = \: ?[/tex]
[tex] \\ \\ [/tex]
Solution:-
we know :-
[tex] \bf \dag \boxed{ \rm work = force \times distance}[/tex]
[tex] \\ \\ [/tex]
So:-
[tex] \dashrightarrow \sf work = force \times distance[/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \sf work =225 \times 10[/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow \bf work =\red{2250\: joule}[/tex]
