Answer to Question #186036 in Mechanics | Relativity for Christina shumet

Question

In a certain machine and effort 120KN acts through 13M to raise a load of 1800KN through a distance of 600cm what is the efficiency of the machine

Answer 1

Explanations & Calculations

• Work performed by the effort here is

$\qquad\qquad \begin{aligned} \small W_e&=\small F\times s=120kN\times 13m\\ &=\small 120\times 13 \,kJ \end{aligned}$

• The effective work received on the system is

$\qquad\qquad \begin{aligned} \small W_l&=\small 1800\,kN\times 0.600m\\ &=\small 1800\times 0.6\,kJ \end{aligned}$

• The efficiency of a machine according to the definition is

$\qquad\qquad \begin{aligned} \small Eff&=\small \frac{\text{Output work (power)}}{\text{Input work(power)}}\times100\\ \end{aligned}$

• Since the effort & the effective work take place during the same period of time(t), the efficiency of the machine would be

$\qquad\qquad \begin{aligned} \small Eff&=\small \frac{1800\times0.6}{120\times 13}\times 100\\ &=\small 69.23 \% \end{aligned}$