Calculating Maximum Acceleration and Speed for a 200kg Load with a 5kW Winch

  • Thread starter Thread starter furor celtica
  • Start date Start date
  • Tags Tags
    Load Winch
AI Thread Summary
The discussion focuses on calculating the maximum acceleration and speed for a 200 kg load using a 5 kW winch. The initial calculation for acceleration was determined to be 2.5 m/s², but there was confusion regarding the gravitational constant used. Participants confirmed that the acceleration value was correct, but there was uncertainty about the calculation for maximum speed. The final speed calculation yielded 2.55 m/s, prompting questions about potential errors in the approach. Overall, the thread highlights the importance of accurately applying physics equations in practical scenarios.
furor celtica
Messages
69
Reaction score
0

Homework Statement



A winch is used to raise a 200 kg load. The maximum power of the winch is 5 kW. Calculate the greatest possible acceleration of the load when its speed is 2 ms^1, and the greatest speed at which the load can be raised.

Homework Equations





The Attempt at a Solution



Alright so this is how I went about things: power=force x velocity, so upward force = 5000/2 = 2500. So net upward force = 500 N, so greatest possible acceleration = 500/200 = 2.5 ms^2.
I know this is incorrect, where did I go wrong?
 
Physics news on Phys.org
Were you expected to use g = 10m/s2, or a more accurate value such as g = 9.81 m/s2 ?
 
just g=10 ms^2
 
furor celtica said:
just g=10 ms^2

Then I don't see anything wrong with your value for the acceleration when velocity is 2 m/s.
 
on another site i got the same answer.
i guess its a typo, thanks for your time
 
alrighty now I've got an issue with ths second problem, finding the greatest speed at which the load can be raised
so greatest speed means the upwards force is equal to the downwards force, so the upwards force will be 2000N
power being 5000, speed is 2.5 ms^1
however the answer is 2.55, another typo or did i truly get it wrong this time?
 
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