Question

An effort of 50N is applied on a simple machine through a distance of 0.6m. As a result, a load of 300N moves through a distance of 7.5 cm.

Calculate the:
(i) Velocity Ratio
(ii) Mechanical Advantage
(iii) Work Done on the machine, and
(iv) Work done by the machine,

Is the work done by the machine equal to the work done on the machine? Explain why.​

Answer 1

Answer:

An effort of 50N is applied on a simple machine through a distance of 0.6m. As a result, a load of 300N moves through a distance of 7.5 cm.

(i) Velocity Ratio  = 8

(ii) Mechanical Advantage = 6

(iii) Work Done on the machine  = 30 J

(iv) Work done by the machine  =  22.5 J

Explanation:

Given,

Effort = 50 N                              Load = 300N

Effort distance = 0.6 m             Load distance = 7.5 cm = 0.075m

• velocity ratio can be calculated by the equation,

$\text { Velocity ratio(VR) }=\frac{\text { Distance moved by the effort }}{\text { Distance moved by the load }}$

                      VR     =   $\frac{0.6}{0.075} = 8$

The mechanical advantage of a machine is given by,

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

                                     M.A    = $\frac{300}{50} = 6$

Efficiency = $\frac{MA}{VR}$ $\times$ 100 %   = $\frac{6}{8} \times$100 =  75%

We know,

${Work }=\mathrm{Force} \times \mathrm{distance}$

or, W   =    F$\times$S

Work done on the machine is called input work ($W_{in}$) and work done by the machine is called output work ($W_{ou}$).

Also Efficiency = $\frac{W_{ou}}{W_{in}}$

• Work Input       =    Effort $\times$Effort distance

                        =     50 $\times$ 0.6 = 30 J

• Work output    =  Load $\times$ Load  distance

                        =  300 $\times$ 0.075 = 22.5 J