Question

A load of 400N is lifted by a first class lever in which the load is at the distance of 20cm and the effort is at the distance of 60cm from the fulcrum. If 150N effort is required to lift the load, what is its efficiency ​

Answer 1

Answer:

efficiency =0.89

which means that the machine is not 100 percent efficient...

Explanation:

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Answer 2

Answer:

A load of 400N is lifted by a first-class lever in which the load is at the distance of 20cm and the effort is at the distance of 60cm from the fulcrum. If 150N effort is required to lift the load its efficiency will be $88.88\%$

Explanation:

• For a lever, the efficiency can be expressed as,

                  $\text { Efficiency}\ (\eta) = \frac{\text { M.A }}{\text { V.R } }$                  ...(1)

• M.A is the mechanical advantage and V.R is the velocity ratio. It can be defined as,

                  $\text{ { Mechanical\ advantage }(M . A)}=\frac{\text { Load }}{\text { Effort }} = \frac{L}{E}$    ...(2)

                  $\text { Velocity ratio }(\text { V.R })=\frac{E_{d}}{L_{d}}$         ...(3)        

         Where,

         L   -    Load

         E   -   Effort

        $L_d$  -   Load distance

        $E_d$  -   Effort distance

In the question, it is given that,

$L = 400\ N$                                 

$E = 150\ N$            

$L_d = 20\ cm=0.20\ m$                  

$E_d =60\ cm = 0.60\ m$  

Substitute the above values into equation (1)

$\text M \cdot \text A=\frac{400}{150}=\frac{40}{15} =2.6666$

V.R  can be obtained by substituting the values of $E_d$ and $L_d$ into equation (2)

$\text V \cdot \text R =\frac{0.60}{0.20}=3$

Using equation (3), Efficiency is,

$\eta = \frac{\frac{40}{15} }{3} \times 100 = 88.88 \%$