Graphs of Work & Power for Falling Stone

  • Thread starter Thread starter leftywefty
  • Start date Start date
  • Tags Tags
    Graphs
AI Thread Summary
The discussion focuses on creating graphs for work and power related to a 5kg stone falling 15m under gravity. The calculated work done by the stone is 735J, but there is confusion about how to plot the graphs correctly. Participants suggest using force and displacement for the work graph and exploring the relationship between power, force, and velocity for the power graph. Additionally, the conversation emphasizes analyzing energy changes, including potential and kinetic energy, over time and distance. Understanding these relationships is crucial for accurately representing the work and power graphs.
leftywefty
Messages
1
Reaction score
0
Draw graphs??

Homework Statement



A stone of a mass of 5kg drops through a distance of 15m under the influence of gravity. Draw graphs of the work done by the stone and the power of the stone as a function of time. Discuss your results.

Homework Equations



Work=(force)(displacement)
Power=(work)/(time) or Power=(Force)(velocity)



The Attempt at a Solution



Ok, so I solved for work
work= (5kg)(9.8m/s^2)(15m)=735J
I don't know what to do now. I would think I need to plot Force on the Y axis, and displacement on the X axis, but I'm not sure.
I'm having the same issue with the second graph
Power=(735J)/(??) The problem says to graph power of the stone as a function of time, so I'm guesing I have to use the power equation that involves time. Can I use Power=(Force)(velocity)? I'm so confused! Help please!
 
Physics news on Phys.org


Look at this problem from the point of view of energy. What can you say about the total energy (potential + kinetic) at all times? How does the potential energy change with distance? With time? (how does distance fallen or y position change with time?). Using this can you determine the change in kinetic energy as a function of distance and time?

AM
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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