What is the work done by a varying force on an object moving along the Y-axis?

[dyinfrmphysic]

Homework Statement

calculate the work done by a force F=(4.00yj)N(a varying force) that acts on an object that moves along the Y-axis, from Y=-2.0m to Y=3.0m

Homework Equations

non given but i assume it is W=fscos(θ)

The Attempt at a Solution

attempted to use W=fscos(θ) but only made me more lost. i have no idea where to even begin

 

[SteamKing]

What's the direction of F as it moves from y = -2 to y = 3?

Is the direction of F the same as the path from y = -2 to y = 3?

 

[dyinfrmphysic]

What's the direction of F as it moves from y = -2 to y = 3?

Is the direction of F the same as the path from y = -2 to y = 3?

I would assume so, because it is acting on the object that is moving, so technically it is the force causing the movement is moving in the same direction. But no other info was given.

 

[clamtrox]

Start from the actual definition of work, W = \int_\gamma d\vec{s} \cdot \vec{F} where you integrate over the path of the object.

 

[dyinfrmphysic]

Start from the actual definition of work, W = \int_\gamma d\vec{s} \cdot \vec{F} where you integrate over the path of the object.

no clue on how to do it this way.

 

[clamtrox]

no clue on how to do it this way.

You know how to integrate and do dot products, right?

You have d\vec{s} = dy \hat{j} (as the object moves on y-axis) and \vec{F} = 4 y N \hat{j}.

 

[HallsofIvy]

I would assume so, because it is acting on the object that is moving, so technically it is the force causing the movement is moving in the same direction. But no other info was given.

You don't have to "assume" so that information is expressly given:
"calculate the work done by a force F=(4.00yj)N(a varying force) that acts on an object that moves along the Y-axis, from Y=-2.0m to Y=3.0m"

That "j" in the force is the unit vector in the x direction. That means that the cos(\theta) in your "W=fscos(θ)" is irrelevant. \theta= 0 so cos(\theta)= 1. But just "W= fs" is wrong because that is for constant force.

Now, have you taken or are you taking a Calculus class? Clearly whoever gave you this problem expects you to know that with a variable force, that "product", fs, becomes an integral: \int f(x)dx.

 

[dyinfrmphysic]

You don't have to "assume" so that information is expressly given:
"calculate the work done by a force F=(4.00yj)N(a varying force) that acts on an object that moves along the Y-axis, from Y=-2.0m to Y=3.0m"

That "j" in the force is the unit vector in the x direction. That means that the cos(\theta) in your "W=fscos(θ)" is irrelevant. \theta= 0 so cos(\theta)= 1. But just "W= fs" is wrong because that is for constant force.

Now, have you taken or are you taking a Calculus class? Clearly whoever gave you this problem expects you to know that with a variable force, that "product", fs, becomes an integral: \int f(x)dx.

i did not take calculus, this is a "stepup problem" according to our professor to get us ready for calc based physics.

 

[vela]

Try using the fact that the work is equal to the area under the curve when you plot the force vs. y.