Question

Which of the following is the minimum amount of work done by a hydraulic lift to raise a 150-kg aluminum block 2.0 m vertically?

a. 300 J
b. 2940 J
c. 1470 J
d. 75 J

Answer 1

d. 75 J. Have a nice day.

Answer 2

Important Formulas:

`w=Fd`

`F=ma`

work(measured in joules) = force(measured in newtons) * distance(measured in meters)

force(measured in newtons) = mass(measured in kg) * acceleration(measured in m/s^2)

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Given:

`m=150kg`

`d=2m`

`a=9.8m/s^2` (acceleration due to gravity)

`w=?`

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Finding force:

`F=ma`

`F=150*9.8`

`F=1470N`

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Finding work:

`w=Fd`

`w=1470*2`

`w=2940J`

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`\fbox{Option B}`