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The maximum mass of a load that can be lifted by the jack and the distance covered are:
m = 160.2 Kg
h = 25 cm
Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.
The parameters given are
= 250
= Area of the small piston = π
= 22/7 x
= 0.5
= ?
= Area of the large piston = π
= π x 1
= 3.14
To calculate the force on the large piston, we will use the below formula
/
=
/
Substitute all the parameters into the equation
250/0.5 = /3.14
= 1570 N
To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law
F = mg
1570 = 9.8m
m = 1570/9.8
m = 160.2 Kg
.(take g=9.81ms^-2)
If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be
/ 0.25
=
/
h
250/0.125 = 1570/3.14h
make h the subject of the formula
6280h = 1570
h = 1570/6280
h = 0.25 m
Therefore, the distance through which the load is lifted is 25 cm
Learn more here: brainly.com/question/13596980