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Mathematics 17 Online
OpenStudy (anonymous):

John is lifting a 1.5 kg toy to a height of 1.2 m above the ground. How much work is done??

OpenStudy (anonymous):

\[\int\limits_{0}^{1.2}1.5(1.2-x)dx=1.5\int\limits_{0}^{1.2}(1.2-x)dx\] \[1.5[1.2x-\frac{x^2}{2}]_0^{1.2}=1.08\]

OpenStudy (anonymous):

Will, if you have a physics book, or pick up any physics book you will see that the formula for work is given by: W=(Force)(distance). So simply this would be w=(1.5)(1.2)=1.80

OpenStudy (anonymous):

Doh. I'm so used to working weight over distance problems. Lagrange is right. They do not distribute the weight as kg/m so it is a straight forward problem.

OpenStudy (anonymous):

W (WORK DONE)= mgh = 1.5*1.2*9.8 =17.1 JOULE LET (g=9.8 M/S^2)

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