Variable force (Ahmed Monir - Ahmed Abdelfattah).pptx

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1. Variable Force Sharkya STEM HIGH SCHOOL Presented By: Ahmed Monir & Ahmed Abdelfattah Samir Abdelfattah 12TH GRADE Supervisor: Mr Wageeh Ezzat 2220008 2220003

2. Definition of work done: When you are exerting a force on an object that causes its position to change in the same direction as that of the force, you are doing work on this object. Work done on an object is made up of two main components: force on and displacement of the object. The displacement of an object must happen along the line of action of the force for the force to do work on the object. Work has units of energy because it is defined as an amount of (transferred) energy, so work usually has units of J (joules).

3. Equation of work done: T he movement of our object to be in the x- direction. Looking at the figure below, we see that the horizontal component Fx of the force F is calculated using the formula: Fx = F cos (0)

4. where is the angle that the force makes with the direction of motion of the object? The work being done on the object is done only by this component of the force that is parallel to the direction of travel of the object, so the work W done on an object moving a distance is acted upon by a force F that makes an angle with the direction of motion of the object is : W = Fs cos (0).

5. Examples of work done Suppose you decide to put all your books and magazines in one wooden box. You place the box on a table, and you pull it using a rope attached to the box, as shown in the figure above. This pull generates a motion of the box that is exactly in the direction of the pull, namely precisely to the right. This means you are doing work on the box! Let us do an example The force applied to the box has the same direction as the direction of motion of the box so work is being done on the box by the force.

6. Examples First example: Suppose that you are exerting a constant force of 250 N, and you manage to drag the box towards you over 2 m The work you exerted on the box doing this is W = Fs = 250 N × 2 m = 500 Nm = 500 J. This means that the work done on the box is W = 500 J.   Second example: Now suppose that after this first pull, you are tired, and your second pull is done with only half the force and the box only moves half the distance. In this case, the work done on the box in the second pull is W = F s = 125 N x 1 m = 125 J.  

7. Third example: In the last situation, we suppose that the box is sliding towards you over ice, and you try to stop it. You end up exerting a small force of F = 10 N on the box because you don't have a lot of traction yourself on the ice, and the box comes to a stop after s = 8m. The important thing to note in this situation is that the work done on the box by you is negative because the force you exerted on the box was opposite the direction of movement of the box. You did W = -10 N x 8 m = -80 J of work on the box.

8. What is Variable Force? It is interesting to know that the forces we encounter daily are primarily variable in nature, defined as a variable force. A force is said to work on a system if there is a displacement in the system when the force is applied in the direction of the force. In the case of a variable force, integration is important to calculate the work done. A work done by a constant force of magnitude F, as we know that displaces an object by Δx , can be given as: W = F.Δx

9. In the variable force, work is calculated with the help of integration. For example, in the case of a spring, the force acting upon any object attached to a horizontal spring can be given as: Fs = - kx k = spring constant x = displacement of the object attached We see that this force is proportional to the displacement of the object from its equilibrium position, so the force acting at each moment during the compression and extension of the spring will be different. To calculate the total work done, one must count the infinitesimal contributions of work done during each moment.

10. Therefore,

11. QUESTION A bullet weighing 40g is moving at a velocity of 600m/s. This bullet strikes a windowpane and passes through it. Now, its velocity is 400m/s. Calculate work done by a bullet when passing through this obstacle. Ans.  K.E = 1/2 (mv 2 1  - mv 2 2 ) Therefore, in this equation, m is 40g or 0.04 kg. Initial velocity (v1) is 600m/s, while final velocity (v2) is 400m/s. (K.E.) = 1/2 ((0.04) (600) 2 −0.04(400) 2 ) K.E = 4000 Joules

12. Consider a graph of variable power vs. displacement, as shown in the figure. Here, the small division represents the displacement Δx due to the force F(x) acting at that point. Assuming that the quantity Δx is small, the force F(x) acting in this duration can be considered a constant force. The area bounded by a rectangle with a length equal to the magnitude of the force F(x) and a width equal to the displacement Δx gives the work done by the force during this duration. Mathematically ,  ΔW =F (x) Δx

13. We notice that the displacements approach zero; the following equation gives the total work done by the force , 

14. Thanks for listening Ahmed Abdelfattah Samir [phone] [email protected] Ahmed Monir Abdelsabour [phone] [email protected] Grade 12 B2 2022 - 2023 Supervisor: Mr Wageeh Ezzat