a) The force that must be applied is 73.5 N
b) The actual efficiency is 82 %
Explanation:
a)
Since the jack is 100% efficient, all the input work is converted into output work. So we can write:
[tex]W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}[/tex]
where:
[tex]F_{in}[/tex] is the input force applied on the jack
[tex]d_{in}[/tex] is the arm of the input force
[tex]F_{out}[/tex] is the output force applied on the jack
[tex]d_{out}[/tex] is the arm of the output force
Here we have:
[tex]F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N[/tex] is the output force (half the weight of the car)
[tex]d_{in} = 40.0 cm = 0.40 m[/tex] is the arm of the input force
[tex]d_{out} = 5.0 mm = 0.005 m[/tex] is the arm of the output force
Solving for [tex]F_{in}[/tex], we find the force that must be applied in input to lift the car:
[tex]F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N[/tex]
b)
The efficiency of the jack is given by the ratio between the output work and the input work:
[tex]\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}[/tex]
where we have:
[tex]F_{out}=5880 N[/tex] is the output force (half the weight of the car)
[tex]d_{in} = 40.0 cm = 0.40 m[/tex] is the arm of the input force
[tex]d_{out} = 5.0 mm = 0.005 m[/tex] is the arm of the output force
Here we are told that the input force this time is
[tex]F_{in}=90.0N[/tex]
Substituting into the equation, we find the new efficiency of the jack:
[tex]\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82[/tex]
Which means an efficiency of 82%.
Learn more about levers:
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