Work energy theorem problem -- Block sliding up a curved incline

[tellmesomething]

Homework Statement

A 10kg small block is pulled in the vertical plane along a frictionless surface in the form of an arc of a circle of radius 10m. The applied force is 200N as shown in the figure. If block is started from rest at A then speed at B would be? (g=10m/s^2)

Relevant Equations

None

I did: Work done by gravity+work done by applied force= KE(final)- KE(initial)
Work done by gravity should simply be -mgh
=100*5=-500J
For work done by applied force we know:
W=∫F⋅ds
which can also be written as
W=∫Fdscos(θ)
since F is constant i can take that out
W=Fcos(θ)∫ds
here since ive resolved the dot product isnt ds the length of the arc?
Following that i get
arc length= \frac{\pi}{3} \times 10
= (3.14*10)/3
Now im stuck here as im not sure what the angle is? Is it the angle between displacement and the force or distance travelled(arc) and the force?
Any help would be appreciated

 

[TSny]

Imagine you are the one doing the work by pulling the string horizontally with the force $\vec F$. When your hand undergoes a small displacement $\vec{ds}$, what is the direction of $\vec{ds}$? What is the value of the angle $\theta$ between your force $\vec F$ and the displacement of the force $\vec ds$?

 

[tellmesomething]

Imagine you are the one doing the work by pulling the string horizontally with the force $\vec F$. When your hand undergoes a small displacement $\vec{ds}$, what is the direction of $\vec{ds}$? What is the value of the angle $\theta$ between your force $\vec F$ and the displacement of the force $\vec ds$?

is it 30 deg?

 

[TSny]

is it 30 deg?

I don't believe so. I interpret the figure as showing a string attached to the block with the string passing over a little pulley. The pulley is fixed in place and the force F is applied at a point on the horizontal part of the string, such as the red dot shown:

The displacement of the red dot corresponds to the displacement of the force. When the string is pulled a little bit, in what direction does the red dot move? How does this direction compare to the direction of the force?

 

[tellmesomething]

I don't believe so. I interpret the figure as showing a string attached to the block with the string passing over a little pulley. The pulley is fixed in place and the force F is applied at a point on the horizontal part of the string, such as the red dot shown:
View attachment 342037

The displacement of the red dot corresponds to the displacement of the force. When the string is pulled a little bit, in what direction does the red dot move? How does this direction compare to the direction of the force?

but i thought it would just be tangential to the arc? isnt that the dirn of displacement in circular motion

 

[TSny]

The infinitesimal displacements of the block are tangent to the arc. But that doesn't enter into calculating the work done on the system by the applied force F.

Think of the system as consisting of the block and the string. Three external forces act on this system: the force of gravity, the normal force acting on the block from the frictionless surface, and the horizontal force applied to the string. According to the work-energy theorem, the change in KE of the system equals the net work done by these three forces. You've already got the work done by the force of gravity. You still need the work done by F and the work done by the normal force.

 

[tellmesomething]

The infinitesimal displacements of the block are tangent to the arc. But that doesn't enter into calculating the work done on the system by the applied force F.

Think of the system as consisting of the block and the string. Three external forces act on this system: the force of gravity, the normal force acting on the block from the frictionless surface, and the horizontal force applied to the string. According to the work-energy theorem, the change in KE of the system equals the net work done by these three forces. You've already got the work done by the force of gravity. You still need the work done by F and the work done by the normal force.

Oh so its just 0?

 

[PeroK]

Oh so its just 0?

Yes.

 

[tellmesomething]

Yes.

should i have calculated the work done by the net force on the block i.e tension minus the component of mg ? instead of just tension or does the work done by gravity already accounts for it?

 

[PeroK]

should i have calculated the work done by the net force on the block i.e tension minus the component of mg ? instead of just tension or does the work done by gravity already accounts for it?

In general, you calculate the work done by each force separately. You should know that by now.

Look at the original diagram and imagine you are applying for $F$ over a certain distance. The work done by that force is independent of what the force is doing inside the box.

Also, the string outside the box and the force $F$ could be at any angle to the fixed pulley and the work done is the same.

 

[tellmesomething]

In general, you calculate the work done by each force separately. You should know that by now.

Look at the original diagram and imagine you are applying for $F$ over a certain distance. The work done by that force is independent of what the force is doing inside the box.

Also, the string outside the box and the force $F$ could be at any angle to the fixed pulley and the work done is the same.

i see ok thankyou.......