Hi, we have applied that in a way. Remember, theta in the formula wd=FsCos(theta) is the angle between the force applied and direction of motion it causes, of which in this case is zero so Cos(0)=1, hence the formula becomes wd= Fs. I.e the force 100N is parallel to the incline plane. Suppose the force applied was horizontal, we would have used 15° as theta in the formula in order to enable us use the resolved force (f=FCos(theta)) which should be parallel to the direction of motion it causes. Please let me know if this response helped 🙏
@@KevinOchola here we consider the work done by the force is on the hypotenuse of the triangle as the displacement caused right? thank you for sharing the explanation
Sir I didn't get the difference between the first and the second question, they seem so similar. And also for the energy output how i know 3000 is input and 3863.7 is output? Any key word is not mentioned as far as I can fetch
Hello! Work input results from the effort force (force applied by the user) for instance, in (b) the force moving the trolley parallel to the inclined plane (AB) working against friction. Work output is work done by the machine (inclined plane) against the weight of the load, which acts vertically, I.e. in (a) work done on the trolley (load) by the inclined plane, is work output . Simply put, work input is due to effort force, while work output is due to the machine against the weight of load. Please let me know if this response helped 🙏
Note that the inclined plane has to help lift the load through the vertical height (h) hence (work output= Weight × h). Also, a force has to be applied to pull the load along plane AB, hence (work input= Force applied × distance AB)
@@KevinOchola Ah yes thank you so much sir, hat's off for being an amazing teacher. I understood now the difference between what the 2 questions were asking. Can you make a video about this topic of inclined planes but with more examples? I have term exams and that would be dope!
Hi! We use sine since we already have the opposite side which makes it easier to obtain the hypotenuse (AB) given angle 15°. Using cosine may not be easy since we don't know both the adjacent and hypotenuse. Please let me know if this response helped 🙏
Hello! mgsintheta is already given as 100N. Since the trolley is in motion with constant velocity (the force applied remains constant throughout, 100N), Resultant force=0. I.e., resultant force = Force applied - (friction+ mgsintheta). But the trolley is assumed frictionless. So the equation becomes: 0= force applied -(0+mgsintheta) Which implies: Force applied=mgsintheta. Please let me know if this response helped 🙏
Note that, mgsintheta is a force acting parallel the inclined plane but opposite the direction of applied force. This force together with frictional force act in the same direction.
Thanku sir.love and respect from india.
Thank you...much appreciated!
i want to ask, why in the second part the formula W = F s cosθ is not used, as here cosθ wasnt used.
please tell
thank you
Hi, we have applied that in a way. Remember, theta in the formula wd=FsCos(theta) is the angle between the force applied and direction of motion it causes, of which in this case is zero so Cos(0)=1, hence the formula becomes wd= Fs. I.e the force 100N is parallel to the incline plane. Suppose the force applied was horizontal, we would have used 15° as theta in the formula in order to enable us use the resolved force (f=FCos(theta)) which should be parallel to the direction of motion it causes.
Please let me know if this response helped 🙏
@@KevinOchola here we consider the work done by the force is on the hypotenuse of the triangle as the displacement caused right?
thank you for sharing the explanation
Sure, you are right 👍
okay thank you @@KevinOchola
Sir I didn't get the difference between the first and the second question, they seem so similar. And also for the energy output how i know 3000 is input and 3863.7 is output? Any key word is not mentioned as far as I can fetch
Hello! Work input results from the effort force (force applied by the user) for instance, in (b) the force moving the trolley parallel to the inclined plane (AB) working against friction.
Work output is work done by the machine (inclined plane) against the weight of the load, which acts vertically, I.e. in (a) work done on the trolley (load) by the inclined plane, is work output . Simply put, work input is due to effort force, while work output is due to the machine against the weight of load. Please let me know if this response helped 🙏
Note that the inclined plane has to help lift the load through the vertical height (h) hence (work output= Weight × h).
Also, a force has to be applied to pull the load along plane AB, hence (work input= Force applied × distance AB)
@@KevinOchola Ah yes thank you so much sir, hat's off for being an amazing teacher. I understood now the difference between what the 2 questions were asking. Can you make a video about this topic of inclined planes but with more examples? I have term exams and that would be dope!
@@KevinOchola Thanks sir!
@zzaenabz it's alright, let me organize.
Why did we use sin instead of cos
Hi! We use sine since we already have the opposite side which makes it easier to obtain the hypotenuse (AB) given angle 15°. Using cosine may not be easy since we don't know both the adjacent and hypotenuse. Please let me know if this response helped 🙏
Why not using the mgsintheta
Hello! mgsintheta is already given as 100N.
Since the trolley is in motion with constant velocity (the force applied remains constant throughout, 100N), Resultant force=0. I.e., resultant force = Force applied - (friction+ mgsintheta).
But the trolley is assumed frictionless. So the equation becomes:
0= force applied -(0+mgsintheta)
Which implies:
Force applied=mgsintheta.
Please let me know if this response helped 🙏
Note that, mgsintheta is a force acting parallel the inclined plane but opposite the direction of applied force. This force together with frictional force act in the same direction.