Pre Calc
posted by Em
Find the work done by a force F of 30 pounds acting in the direction (2,3) in moving an object 3 feet from (0,0) to the point in the first quadrant along the line y = (1/2)x.
The answer is 85.38 footpounds.
I just need to figure out how to get the answer.

Damon
Find the angle of the Force F
cos T = 2/sqrt13 = .555
T = 56.3 deg above x axis
or
tan T = 2/3
so T = 56.3
Find the direction of motion from slope of line
tan slope = 1/2
slope angle = 26.6 deg
angle between force and motion = 56.326.6 = 29.7 degrees
so
component of force in direction of motion = 30 cos 29.7
= 26.06 pounds
26.06*3 = 78.2 ft lbs
beats me how to get 85.38
It would be much closer if the force vector were direction(3,2) 
Megha
the vector is actually <2,2>
and the line angle is 64.3 
Oscar
the vector is indeed <2,2>, but the line angle (angle formed by y=1/2x and xaxis) is as Damon posted, 26.6 degree. As for why it isn't 64.3 degree is because slope is change in y over the change in x, thus this means horizontal change is 2 and vertical change is 1. When finding the line angle, simply use arc tan(opp/adj) or in this case arc tan(1/2) which will result in 26.6 degree.
As for the answer, its actually 95.36 ft lb if the vector is <2,2> 
Katelyn
F1 > 30 lbs, direction <2, 2>
magnitude of <2, 2> = 2sqrt(2)
Find unit vector of F1 (divide original components by magnitude)
> <30/sqrt(2), 30/sqrt(2)>
F2 > 3 feet, direction y = 1/2x
3 = sqrt(x^2 + y^2)
y = 1/2x
3 = sqrt(x^2 + (1/2x)^2)
9 = 5/4* x^2
x^2 = 36/5
x = 6/sqrt(5)
y = 2/sqrt(5)
F1 = <30/sqrt(2), 30/sqrt(2)>
F2 = <6/sqrt(5), 3/sqrt(5)>
Find dot product of F1 and F2
(30/sqrt(2))*(6/sqrt(5))+(30/sqrt(2))*(3/sqrt(5)) = 85.38
Answer in the back of the book is 85.38
Respond to this Question
Similar Questions

multivariable calc with some high school physics
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 1 meters. What is the magnitude of the tension on the ends of the clothesline? 
physics
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a positive magnitude of 6.7N; a second force has a magnitude of 4.6 N and points in the negative … 
physics
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a positive magnitude of 6.7N; a second force has a magnitude of 4.6 N and points in the negative … 
physics
If there is just one force acting on an object, does its work necessarily result in an increase in kinetic energy? 
Pre Calc
Find the work done by a force F of 30 pounds acting in the direction (2,3) in moving an object 3 feet from (0,0) to the point in the first quadrant along the line y = (1/2)x. 
physics
A)An object at rest has no forces acting on it. B)An object can be moving in one direction while the net force acting on it is in another direction. C)In order to move a massive crate sitting on the floor, the force you apply to the … 
Physics
A two dimensional object placed in the xy plane has three forces acting on it: a force of 3N along the x axis acting at a point (3m, 4m); A force of 2N along the y axis acting at (2, 5m); and a force of 5N in the negative x direction … 
PHYSICS
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.2 N ; a second force has a magnitude of 5.0 N and points in the negative y direction … 
calculus
A variable force of 7xâˆ’2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 19 ft. (Round your answer to two decimal places.) 
PreCalculus Help?
A 5lb 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 footpound. a. 11 ft. * lbs b. 34 ft. * lbs c. 56 ft. * lbs d. …