Question

A force of 4 pound acts in the direction of 48degree to the horizontal. The force moves an object along a straight line from the point (1,7) to the point (8,10), with distance measured in feet. What is the work done?

Answer 1

Answer:

Work Done = 27.66 Joules

Step-by-step explanation:

The work done is:

$W=F*d*Cos \theta$

Where

F is the fore

d is the distance moved

$\theta$  is the angle

The force is "4"

The distance is the distance from (1,7) to (8,10)

The distance is the square root of change in y coordinates, squared and sum of change in x coordinates, squared. Hence, distance is:

$\sqrt{(10-7)^2+(8-1)^2}\\=\sqrt{3^2+7^2}\\\sqrt{58}$

The angle is gotten by the formula:

$Cos\phi = \frac{Force in x direction}{Distance}\\Cos\phi = \frac{8-1}{\sqrt{58}}\\Cos\phi=\frac{7}{\sqrt{58}}\\\phi = Cos^{-1}\frac{7}{\sqrt{58}}\\\phi = 23.21$

So, now:

$\theta = 48 - 23.21 = 24.79$

Work Done = $F*d * Cos\theta = 4 * \sqrt{58} * Cos(24.79)$

Work Done = 27.66 Joules