Question

12. Suppose a person uses a mechanical iack to lift half the weight of a car with a

Answer 1

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:

$W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}$

where:

$F_{in}$ is the input force applied on the jack

$d_{in}$ is the arm of the input force

$F_{out}$ is the output force applied on the jack

$d_{out}$ is the arm of the output force

Here we have:

$F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Solving for $F_{in}$, we find the force that must be applied in input to lift the car:

$F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N$

b)

The efficiency of the jack is given by the ratio between the output work and the input work:

$\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}$

where we have:

$F_{out}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Here we are told that the input force this time is

$F_{in}=90.0N$

Substituting into the equation, we find the new efficiency of the jack:

$\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82$

Which means an efficiency of 82%.

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