Answer to Question #185979 in Mechanics | Relativity for buddy

Question

A force of 577 N acts for 3.99 m over 4.55 s. That force is the product of 708 J of energy to the system that is supplied for 1.20 s. What is the efficiency of the system?

Answer 1

Explanations & Calculations

• The efficiency is defined as

$\qquad\qquad \begin{aligned} \small Eff&=\small \frac{Work \,\,output}{Work\,\, input}\times 100 \end{aligned}$

• Dividing both numerator and denominator yields another form for efficiency with power.

$\qquad\qquad \begin{aligned} \small Eff&=\small \frac{Work\,output/t}{Work \, input/t}\times100\\ &=\small \frac{Power \,output}{Power \,input}\times 100 \end{aligned}$

• When power is given in each situation, this form can be used

• Input power is

$\qquad\qquad \begin{aligned} \small &=\small \frac{708J}{1.20s}=590W \end{aligned}$

• Power output from the system is

$\qquad\qquad \begin{aligned} \small &=\small \frac{work}{time}=\frac{577N\times 3.99m}{4.55s}=505.98 W \end{aligned}$

• Then,

$\qquad\qquad \begin{aligned} \small Eff&=\small \frac{505.98\,W}{590\,W}\times 100\\ &=\small \bold{85.76\%} \end{aligned}$