for these problems can we assume that the system is at equilibrium so we set the net forces equal to zero? Then whatever we get for net force we multiply it by the cosine of the incline times the distance traveled to get the net work done?
If you still define "up the ramp" as positive, then Fgx would still be negative (since it is down the ramp.) Work, however, would end up being negative since both Fgx and displacement are both negative (down the ramp.) One last thing to remember if you are given this problem is that the direction of friction would now be up the ramp to oppose the motion of the block coming down.
In this problem the applied force is parallel to the surface, so there is no component into the surface. However, if you pulled up at an angle to the surface, then the Fn would be mgcos(theta) - Fsin(angle). Here is another video that covers this concept. th-cam.com/video/ipJVYnLMWVo/w-d-xo.html
Not exactly, but I do have one that finds the force. To find the work, simply multiply by the distance traveled. th-cam.com/video/ipJVYnLMWVo/w-d-xo.html
Just to clarify, why did the friction work use cos(30°) and the applied work (400J) did not? If you rotated the FBD by 30° clockwise, both forces lie on the x-axis and do not need cos(30°)…? (I think the net work is around 228.325J)
Force of Friction is (mu)x(normal force). Normal force can be found from the component of gravity into the inclined plane which is mgcos(theta). Here is a video with more details. th-cam.com/video/hTuBIjG-Q98/w-d-xo.html Thanks for the question!
80N force was already parallel to the plane of the surface, so no need to "fix it." If it had been drawn at an angle to the surface, then you would need to find the parallel component. See this video for an example of this : th-cam.com/video/ipJVYnLMWVo/w-d-xo.html
Good question. Work is equal to the change in energy. In this problem, the object is moving at a constant velocity, so that the KE is not changing. All the work is turning into gravitational PE and thermal energy from friction. Technically, there must be a bit of extra work to get the object moving in the first place. Though, this would be cancelled out by the negative work when it stops at the top. Practically though, the crux of the problem occurs during the constant velocity phase and KE is ignored in the problem.
at 5:58 in the video: (0.2)(5)(9.8)cos(30degrees)(5) is equal to 24.5 J not 42.4 J, 42.4 J would not account for cos(30) which is (0.5); therefore the net work on the object is equal to 253 J not 235 J
Everything is the same regardless of the initial velocity. In fact, the net work can be used to find the change in velocity since Net work = change in kinetic energy.
Thank you so much! My Ivy League professor couldn’t teach this if his life depended on it! Good thing this 7 minute video helped explain this incline problem! Do you also make videos on string theory I have always wanted to be an physics major but I’m not sure if I need to be good at string theory to do so!??? Thanks you so much!!!
An object of mass m=1kg is sliding from top to bottom in the frictionless incline plane of inclination angle 0=30o and the length of inclined plane is 10m as shown in the figure.Calculate the work done.
for these problems can we assume that the system is at equilibrium so we set the net forces equal to zero? Then whatever we get for net force we multiply it by the cosine of the incline times the distance traveled to get the net work done?
Thank you❤ understood briefly, love from India 🇮🇳
Thank you for helping me before my exam today
What if the object was sliding down the ramp instead of up? would Fgx be negative or positive?
If you still define "up the ramp" as positive, then Fgx would still be negative (since it is down the ramp.) Work, however, would end up being negative since both Fgx and displacement are both negative (down the ramp.) One last thing to remember if you are given this problem is that the direction of friction would now be up the ramp to oppose the motion of the block coming down.
is there not a component of the applied force that would make Fn be mgcostheta - Fsintheta?
In this problem the applied force is parallel to the surface, so there is no component into the surface. However, if you pulled up at an angle to the surface, then the Fn would be mgcos(theta) - Fsin(angle). Here is another video that covers this concept. th-cam.com/video/ipJVYnLMWVo/w-d-xo.html
Great video, Can you please make up a question with the applied force at an angle to the ramp asking for the net work?
Not exactly, but I do have one that finds the force. To find the work, simply multiply by the distance traveled.
th-cam.com/video/ipJVYnLMWVo/w-d-xo.html
for the net work, do we supposed to + W from Normal force also?
No. SInce the normal force is perpendicular to the distance traveled, it does not do any work on the block.
Just to clarify, why did the friction work use cos(30°) and the applied work (400J) did not?
If you rotated the FBD by 30° clockwise, both forces lie on the x-axis and do not need cos(30°)…? (I think the net work is around 228.325J)
Force of Friction is (mu)x(normal force).
Normal force can be found from the component of gravity into the inclined plane which is mgcos(theta).
Here is a video with more details.
th-cam.com/video/hTuBIjG-Q98/w-d-xo.html
Thanks for the question!
if the object was move down the plane do we still do find the gravitational work also?
Yes you could. Work due to gravity would be the same.
Why you did not put sin 30 in the 80n force?
80N force was already parallel to the plane of the surface, so no need to "fix it." If it had been drawn at an angle to the surface, then you would need to find the parallel component. See this video for an example of this : th-cam.com/video/ipJVYnLMWVo/w-d-xo.html
If the object is moving, why isn’t there any kinetic energy as part of work? 😢
Good question. Work is equal to the change in energy. In this problem, the object is moving at a constant velocity, so that the KE is not changing. All the work is turning into gravitational PE and thermal energy from friction.
Technically, there must be a bit of extra work to get the object moving in the first place. Though, this would be cancelled out by the negative work when it stops at the top. Practically though, the crux of the problem occurs during the constant velocity phase and KE is ignored in the problem.
Thank you Very helpfull
at 5:58 in the video: (0.2)(5)(9.8)cos(30degrees)(5) is equal to 24.5 J not 42.4 J, 42.4 J would not account for cos(30) which is (0.5); therefore the net work on the object is equal to 253 J not 235 J
Cos(30) is (0.866)
Sin(30) is (0.5)
What if there was velocity?
Everything is the same regardless of the initial velocity. In fact, the net work can be used to find the change in velocity since Net work = change in kinetic energy.
Character In the video It's great, I like it a lot $$
Thank you so much! My Ivy League professor couldn’t teach this if his life depended on it! Good thing this 7 minute video helped explain this incline problem! Do you also make videos on string theory I have always wanted to be an physics major but I’m not sure if I need to be good at string theory to do so!??? Thanks you so much!!!
Help me
3. If A•B=|A×B| where A and B are two none_ zero vectors find angle between A and B
how about work done due to gravity and normal force
thankyuo brother
Is net work also the change in Kinetic Energy?
Yes
is the 80N force applied a non conservative force?
In this case should be conservative. Work done by friction would be non-conservative.
An object of mass m=1kg is sliding from top to bottom in the frictionless incline
plane of inclination angle 0=30o and the length of inclined plane is 10m as shown
in the figure.Calculate the work done.
i hope you solved this already but the answer should be W = mg * sin(30) * distance, or 1 * 9.8 * sin(30) * 10
@@Azhix yeah...thanks for the reply hihi
How do they decide to use cosine or sine? It always seems so arbitrary
Thanks!
Also check this video for the derivation wrt inclined planes.
th-cam.com/video/hTuBIjG-Q98/w-d-xo.html
Around the 10:30 mark
thankyuo brother
Your very welcome