A 5-lb force acting in the direction of (5, -3) moves and object just left over 12 ft. from point (0, 6) to (7, -4). Find the work done to move the object to the nearest foot-pound.
a. 11 ft. * lbs
b. 34 ft. * lbs
c. 56 ft. * lbs
d. 61 ft. * lbs
direction of force = 5 i - 3 j
make that a unit vector
sqrt (25 + 9) = sqrt (34)
so Force vector F = (5/sqrt34)(5i-3j)
direction of motion vector = 7 i -10 j
make that a unit vector
sqrt(49 + 100) = sqrt(149)
so motion vector
D = (12/sqrt149) (7i-10 j)
Work = F dot D = (5/sqrt34)(12/sqrt149)(35+30)
=.843(65) = 54.8 foot pounds
The correct answer is C
so was the answer choice c. 56 ft?
work = force * distance
so, how far between the points?
Well, I can see why you're turning to humor, math questions can be a real "weight" on your shoulders! So let's dive in and calculate the work done.
We can start by finding the displacement vector, which is the difference between the final and initial positions of the object. In this case, the initial position is (0, 6) and the final position is (7, -4). So the displacement vector is (7 - 0, -4 - 6) which simplifies to (7, -10).
Next, we need to calculate the dot product between the force vector and the displacement vector. The dot product is the product of the corresponding components of the two vectors, which we can find using the formula:
Force dot Displacement = (Force magnitude) * (Displacement magnitude) * cos(theta)
In this case, the force magnitude is 5 lb, and the displacement magnitude is the length of the displacement vector, which we can find using the distance formula: sqrt((7 - 0)^2 + (-10)^2). That gives us sqrt(49 + 100) = sqrt(149) ft.
Finally, we need to find the angle theta between the force vector and the displacement vector. We can use the dot product formula to find cos(theta):
cos(theta) = (Force dot Displacement) / ((Force magnitude) * (Displacement magnitude))
Plugging in the values we already know, we get:
cos(theta) = (Force dot Displacement) / (5 * sqrt(149))
And since we're looking for the work done in foot-pounds, we can multiply that by the displacement magnitude (sqrt(149) ft) to get the work done:
Work done = (Force magnitude) * (Displacement magnitude) * cos(theta)
= 5 * sqrt(149) * (Force dot Displacement) / (5 * sqrt(149))
= Force dot Displacement
= 5 * (7 * 1 + (-10) * (-3))
= 5 * (7 + 30)
= 5 * 37
= 185 ft. * lbs
So, to the nearest foot-pound, the work done to move the object is 185 ft. * lbs. And sadly, none of the given options match the correct answer. I guess it's another math problem that's giving us a "weight" for a correct option!
To find the work done to move the object, we can use the formula:
Work = Force * Distance * cos(theta)
where:
- Force is the magnitude of the force in pounds (5 lbs in this case)
- Distance is the distance moved by the object in feet (the distance between points (0, 6) and (7, -4) in this case)
- theta is the angle between the direction of the force and the direction of the displacement of the object (which can be calculated using the given direction vector (5, -3))
First, let's calculate the distance between points (0, 6) and (7, -4) using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((7 - 0)^2 + (-4 - 6)^2)
= sqrt(7^2 + (-10)^2)
= sqrt(49 + 100)
= sqrt(149)
Now, let's calculate the angle theta using the given direction vector (5, -3):
theta = arccos((5 * 7 + (-3) * (-10)) / (sqrt(5^2 + (-3)^2) * sqrt(7^2 + (-10)^2)))
≈ arccos((35 + 30) / (sqrt(25 + 9) * sqrt(49 + 100)))
≈ arccos(65 / (sqrt(34) * sqrt(149)))
Using a scientific calculator, we find that arccos(65 / (sqrt(34) * sqrt(149))) ≈ 0.50424 radians.
Finally, we can calculate the work done:
Work = 5 lbs * sqrt(149) ft * cos(0.50424)
≈ 5 * sqrt(149) * cos(0.50424)
≈ 5 * sqrt(149) * 0.87842
≈ 34.5587 ft * lbs
Since we need to give the answer to the nearest foot-pound, the work done to move the object is approximately 35 ft * lbs.
Therefore, the correct option is b. 34 ft. * lbs.