The efficiency of the machine is 0.53 (53%).
Explanation:
The efficiency of a machine is given by:
[tex]\eta = \frac{W_{out}}{W_{in}}[/tex]
where
[tex]W_{out}[/tex] is the output work
[tex]W_{in}[/tex] is the work in input
The output work is:
[tex]W_{out}=F_L d_L = (2000)(2)=4000 J[/tex]
where
[tex]F_L = 2000 N[/tex] is the load
[tex]d_L = 2 m[/tex] is the distance covered by the load
The input work is:
[tex]W_{in}=F_E d_E = (1250)(6)=7500 J[/tex]
where
[tex]F_E = 1250 N[/tex] is the effort force
[tex]d_E = 6 m[/tex] is the distance from the effort to the pivot
Solving for the efficiency,
[tex]\eta=\frac{4000}{7500}=0.53[/tex]
So, the efficiency of the machine is 0.53 (53%).
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
#LearnwithBrainly
