Question

A 4-lb. force acting in the direction of (vector) 4,-2 moves an object just over 7ft from point (0,4) to (5,-1). Find the work done to move the object to the nearest foot-pound.

A) 35 ft*lbs
B) 27 ft*lbs
C) 18 ft*lbs
D) 7 ft*lbs

Answer 1

To solve this problem, we have to find the net displacement and the net force and the multiply the dot product together and get the work done.

The work done on moving the object is 27ft*lbs

Work done in moving the object from point A to point B

To find the work done on this object, let's find the net force on the object.

Data;

• force = 4lb
• direction = 4, -2
• displacement = 7ft
• direction = (0, 4) to (5,1)

The unit vector of the force is

$\sqrt{4^2 +(-2)^2} =\sqrt{16 + 4} = \sqrt{20}$

$\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} }$

The net force acting on the object is

$F = 4(\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} })\\F= (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } )$

The displacement on the object is 7ft through (0,4) to (5, -1)

The unit vector on displacement is

$\sqrt{5^2 + (-1-4)^2} = \sqrt{25+25} = \sqrt{50}$

$\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }$

The net displacement will be

$7(\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }) = \frac{35}{\sqrt{50} }, \frac{-35}{50}$

The work done will be F.d

$w = f. d$

$w = (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } ) * \frac{35}{\sqrt{50} }, \frac{-35}{50}\\w = 17.71+ 8.854\\w = 26.567 = 27ft*lbs$

The work done on moving the object is 27ft*lbs

Learn more on work done on an object here;

https://brainly.com/question/26152883

#SPJ1

Answer 2

Answer: the answer is B 27ft

Step-by-step explanation: its cam btw