Shouldn't the magnetic field and current be reverse in the inductor animation at the end of the video? As the current increasing in strength, the magnetic field would proportionally increase in the other direction.
How do you please solve part (e)? Object Sliding Down an Inclined Plane problem: An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an angle of θ = 30 degrees to the ground. The coefficient of kinetic friction µk = 0.2. At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction µk = 0.3 until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface. (a) What is the work done by the friction force while the mass is sliding down the inclined plane? (Is it positive or negative?) (b) What is the work done by the gravitational force while the mass is sliding down the inclined plane? (Is it positive or negative?) (c) What is the kinetic energy of the mass when it just reaches the bottom of the inclined plane? (d) Symbolically, what is the work done by the friction force while the mass is sliding along the ground? Is this positive or negative? Express you answer in terms of some or all of the following: m, µk, g, and d where d is the distance it takes the object to stop measured from the bottom of the incline. (e) How far from the bottom of the inclined plane does the object slide along the rough surface? Thanks!
Since the object starts at rest and ends at rest, the net work done on the object needs to be zero, by the time it reaches its stopping point. The work done by gravity (i.e. its drop in GPE), will therefore equal the total of the heat lost to friction on the incline and on the flat path. The normal force changes from the incline to the flat path, so this will need to be accounted for, in both events. Once you account for the net work adding up to zero, you solve for the distance associated with the work done on friction along the flat path. Everybody brings m*g to the party. The solution I get is: d = L*(sin(theta) - mu_k1*cos(theta))/mu_k2 d = 3.3 meters
Pythagoras strikes again... Who in the world would have thought that the ratio between the sides of a triangle and their hypotenuse would have anything to do with LC circuits 😂😂😂
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I just can say BEAUTIFUL
Finally a video on this that makes sense!
this video popped up the day before i have a test on this exact topic, thank you!
Best of luck on your test!
Shouldn't the magnetic field and current be reverse in the inductor animation at the end of the video? As the current increasing in strength, the magnetic field would proportionally increase in the other direction.
This is a great video! thank you
How do you please solve part (e)?
Object Sliding Down an Inclined Plane problem:
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of
length l = 3.0 m. The plane is inclined by an angle of θ = 30 degrees to the ground. The coefficient of kinetic friction µk = 0.2. At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction µk = 0.3 until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface.
(a) What is the work done by the friction force while the mass is sliding down the
inclined plane? (Is it positive or negative?)
(b) What is the work done by the gravitational force while the mass is sliding down
the inclined plane? (Is it positive or negative?)
(c) What is the kinetic energy of the mass when it just reaches the bottom of the
inclined plane?
(d) Symbolically, what is the work done by the friction force while the mass is sliding
along the ground? Is this positive or negative? Express you answer in terms of
some or all of the following: m, µk, g, and d where d is the distance it takes the
object to stop measured from the bottom of the incline.
(e) How far from the bottom of the inclined plane does the object slide along the
rough surface?
Thanks!
Since the object starts at rest and ends at rest, the net work done on the object needs to be zero, by the time it reaches its stopping point. The work done by gravity (i.e. its drop in GPE), will therefore equal the total of the heat lost to friction on the incline and on the flat path. The normal force changes from the incline to the flat path, so this will need to be accounted for, in both events. Once you account for the net work adding up to zero, you solve for the distance associated with the work done on friction along the flat path. Everybody brings m*g to the party.
The solution I get is:
d = L*(sin(theta) - mu_k1*cos(theta))/mu_k2
d = 3.3 meters
Pythagoras strikes again... Who in the world would have thought that the ratio between the sides of a triangle and their hypotenuse would have anything to do with LC circuits 😂😂😂
🙇🏻♂️
Ah we can dance... if we want to!
Yes all looks good!
I told you I would have to see what mr.p felt like doing in the moment. And that was it!
Thanks for looking this one over.