Given force equation, solve for work.

[bfusco]

Homework Statement

Assume that a force acting on an object is given by , where the constants = 2.9 and = 3.9 . Determine the work done on the object by this force as it moves in a straight line from the origin to = (10.0 + 21.5).

The Attempt at a Solution

well... what i did first was i found the length of the tangent of the distance. (using the pythagorean theorem, to get 23.7m).
i don't really know what to do with the other given information. i thought maybe i could plug the given constants into the equation graph and then find the length of the tangent of the force (which i got to be 4.86 N) and then just plug into the equation W=FΔx, but that is incorrect.

apparently the answer is roughly 1000 J, but I am obviously not getting that as my answer.

 

[emailanmol]

Hey,i think you missed writing the equation. the entire problem has not been provided

 

[bfusco]

Hey,i think you missed writing the equation. the entire problem has not been provided

hahahaha...oops sorry the force is given by the equation F=axi+byj (i, j being the components of the force i (hat) and j (hat)). the constants a=2.9 N/m and b=3.9 N/m

 

[emailanmol]

Hey.

See break force and displacement along the axis.
Force along X axis is given by ax (i)
Remember this force is not a constant and depends on the position of particle on x-axis given by x.

Displacement along x in moving from x to x +dx is given by dx (i)
which makes dW=ax(dx).
Integrate to get workDo similarly for y.

What do you see?

 

[asteriatic]

I would approach this problem by first understanding the given information and the force equation. The force equation, F(x) = kx, represents a linear relationship between force and distance, where k is a constant. In this case, the constants given are k = 2.9 and x = 3.9. This means that the force acting on the object increases by 2.9 N for every 3.9 m it moves.

To solve for work, we can use the formula W = F(x)Δx, where Δx is the distance the object moves. In this case, we are given the starting point (origin) and the end point (10.0 + 21.5), so we can calculate Δx as 10.0 + 21.5 = 31.5 m.

Now, we can plug in the given constants into the force equation to get F(x) = 2.9(3.9) = 11.31 N. Plugging this into the work equation, we get W = (11.31 N)(31.5 m) = 356.865 J. However, this is the work done by the force over the entire distance of 31.5 m.

To find the work done specifically from the origin to the end point, we can use the area under the force-distance curve. Since the force equation represents a straight line, we can calculate this area as a triangle. The base of the triangle is the distance traveled (31.5 m) and the height is the force at the end point (11.31 N). Therefore, the work done from the origin to the end point is (1/2)(31.5 m)(11.31 N) = 178.4325 J.

In order to get the answer of roughly 1000 J, we would need to know the force equation for the entire distance from the origin to the end point, as well as the distance traveled at each point. This would allow us to calculate the area under the curve and get a more accurate result. However, based on the given information and the force equation given, the work done on the object from the origin to the end point is approximately 178 J.