Work Done by a General Variable Force

AI Thread Summary
To calculate the net work done on an 8 kg brick moving along the x-axis with a variable force, one method involves finding the area under the acceleration vs. position graph. Initially, there was confusion about the calculation, but it was resolved by correcting an input error. Another approach discussed involves using the slope of the acceleration graph, which relates to the force acting on the brick. This method simplifies the calculation by directly connecting the slope to the force without needing to find the area. Understanding both methods provides a comprehensive view of solving work done by a variable force.
musicfairy
Messages
101
Reaction score
0
A 8 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Figure 7-37. What is the net work performed on the brick by the force causing the acceleration as the brick moves from x = 0 to x = 8.0 m?

07_34.gif


What I tried to do is find the area and multiply by the mass, but it doesn't work.

Please help.


Edit: Never mind it worked. I just typed it in wrong. But I have another concern.

Is there any additional ways to solve this problem other than the one I mentioned? I was looking at someone else's work and that person did something with the slope. How does that work?
 
Last edited:
Physics news on Phys.org
By looking at it I could see that if they take info from the slope they would appear to be taking a longer route, finding the slope just cancels the displacements.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top