Question

CALCULATE THE WORK AND POWER (image below)

Answer 1

(1) -
`W=20,000 \ J`

(2) -
`W=8.00 * 10^6 \ J`

(3) -
`W=7500\ J` for part a and
`P=375 \ W` for part b

Conceptual:

What is work?

- Work is simply energy transferred from one place to another. Work is a scalar quantity and is measured in Joules, J.

`\boxed{\left\begin{array}{ccc}\text{\underline{Formulas used to calculate work:}}\\\\W=F\Delta r\cos\theta \ \text{(Constant} \ \vec F)\\\\W=\int\limits^(r_2)_(r_1) {F\cos\theta} \, dr \ \text{(Varible} \ \vec F)\end{array}\right}`

** "F" is the applied force on the object, "Δr" is the displacement of the object, "θ" is the measured angle between the applied force and direction of displacement. For our purposes we are assuming θ is 0, which would make the cosine equal 1.**

What is power?

- Power is the measure of work being done per second, J/s. A joule per second is a Watt, W.

`\boxed{\left\begin{array}{ccc}\text{\underline{Formula used to calculate power:}}\\\\P=(W)/(t) \end{array}\right}`

Step-by-step:

Question (1):

Given:

`F= 1000 \ N\\\\\Delta r=20 \ m\\\\Eqn\rightarrow W=F \Delta r`

Plug the known values into the equation.

`\Longrightarrow W=(1000)(20)\\\\\therefore \boxed{\boxed{W=20,000 \ J}}`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Question (2):

Given:

`F= 8000 \ N\\\\\Delta r=1000 \ m\\\\Eqn\rightarrow W=F \Delta r`

Plug the known values into the equation.

`\Longrightarrow W=(8000)(1000)\\\\\therefore \boxed{\boxed{W=8.00 * 10^6 \ J}}`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Question (3):

Given (part a):

`F= 500 \ N\\\\\Delta r=15 \ m\\\\Eqn\rightarrow W=F \Delta r`

Plug the known values into the equation.

`\Longrightarrow W=(500)(15)\\\\\therefore \boxed{\boxed{W=7500\ J}}`

Given (part b):

`W= 7500 \ J\\\\t=20 \ s\\\\Eqn\rightarrow P=(W)/(t)`

Plug the known values into the equation.

`\Longrightarrow P=(7500)/(20) \\\\\therefore \boxed{\boxed{P=375 \ W}}`

Thus, all given questions are solved.