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?
Not exactly, but I do have one that finds the force. To find the work, simply multiply by the distance traveled. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ipJVYnLMWVo.html
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.
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.
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. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hTuBIjG-Q98.html Thanks for the question!
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. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ipJVYnLMWVo.html
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!!!
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 : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ipJVYnLMWVo.html
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.
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
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.
Thanks! Also check this video for the derivation wrt inclined planes. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hTuBIjG-Q98.html Around the 10:30 mark
