Clarification on this specific problem (work done on incline plane)

In summary, the problem considers a block of weight on an inclined plane with an angle \theta and a coefficient of kinetic friction \mu. The task is to find the work done by an applied force, given the variables \mu,w,\theta, L. The solution assumes that the block is not moving, resulting in an equation where the applied force is equal to the sum of the friction force and the weight force. This allows for a simplified solution that only includes the given variables.
  • #1
elsternj
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Homework Statement


A block of weight sits on a plane inclined at an angle [tex]\theta[/tex] as shown. (Intro 1 figure) The coefficient of kinetic friction between the plane and the block is [tex]\mu[/tex].

What is the work done by the applied force of magnitude ?
Express your answer in terms of some or all of the following: [tex]\mu[/tex],w,[tex]\theta[/tex], L

2994_a.jpg

Homework Equations


The Attempt at a Solution


now the answer to this is W = w(sin([tex]\theta[/tex])+[tex]\mu[/tex]cos([tex]\theta[/tex]))L

now I somewhat understand this because the force has to be enough to overcome friction and the weight pulling it down. but what I am concerned with is this:

This would be the answer if the sum of the forces were equal to 0 right?

because if:
[tex]\sum[/tex]Fx=F-wsin[tex]\theta[/tex]-[tex]\mu[/tex]wcos[tex]\theta[/tex]=0

when you solve for F you get the answer originally posted and then just multiply it by the length. How is one to know, given the above information, to set that equal to 0 and solve? It doesn't mention anything about moving at a constant speed. Just trying to get some clarification because I always set the sum of my forces equal to ma and I had trouble answering this question because the answer had to only include the variables [tex]\mu[/tex],w,[tex]\theta[/tex], L ... so the only way I could see to solve for F and keep only those variables would be if acceleration happened to be 0.

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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  • #2
hi elsternj! :smile:
elsternj said:
A block of weight sits on a plane inclined at an angle [tex]\theta[/tex] as shown. (Intro 1 figure) The coefficient of kinetic friction between the plane and the block is [tex]\mu[/tex].

What is the work done by the applied force of magnitude ?
Express your answer in terms of some or all of the following: [tex]\mu[/tex],w,[tex]\theta[/tex], L

It doesn't mention anything about moving at a constant speed.
L ... so the only way I could see to solve for F and keep only those variables would be if acceleration happened to be 0.

yes, you're right, it's a rubbish question :redface:
 

1. What is an incline plane?

An incline plane is a flat, sloped surface that allows for an object to be moved up or down with less force than if it were being moved vertically.

2. How is work done on an incline plane?

Work is done on an incline plane when a force is exerted on an object that is moving along the plane. This force must be parallel to the incline plane, and the work done is equal to the magnitude of the force multiplied by the displacement of the object along the incline.

3. Does the angle of inclination affect the work done on an incline plane?

Yes, the angle of inclination does affect the work done on an incline plane. As the angle increases, the force required to move an object up the incline also increases, resulting in more work being done.

4. What is the relationship between work and energy on an incline plane?

Work and energy are directly related on an incline plane. The work done on an object is equal to the change in its kinetic energy, which is the energy of motion. This means that the work done on an object on an incline plane will result in a change in its speed or direction of motion.

5. How can I calculate the work done on an incline plane?

The work done on an incline plane can be calculated using the formula W = Fd cosθ, where W is work, F is the applied force, d is the displacement along the incline, and θ is the angle of inclination. Alternatively, the work can also be calculated using the change in potential energy of the object, W = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the vertical height of the incline.

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