How much work is done by the crane in lifting the load?

  • Thread starter Thread starter Sirsh
  • Start date Start date
  • Tags Tags
    Building Crane
AI Thread Summary
The discussion revolves around calculating the work done by a crane lifting an 800 kg load to a height of 90 m at a constant speed. The potential energy (PE) is calculated using the formula E = mgh, resulting in 705,600 J. There is a debate about whether to include kinetic energy (KE) in the total work done, with one participant asserting that both PE and KE should be considered due to the conservation of mechanical energy. Another participant questions the formula for work, suggesting a different approach based on force and distance. The conversation highlights the importance of understanding energy concepts in physics calculations.
Sirsh
Messages
262
Reaction score
10
Hi, i was just wondering with this question because I'm unsure if I am supposed to add both kinetic and gravitation energy into this answer? thanks.

1 A crane on a building site lifts an 8.00 × 102 kg load from ground level to a vertical height of 90.0 m at a constant speed of 2.00 m s–1. Ignore the mass of the cable.

A. How much work is done by the crane in lifting the load through this distance?
m = 800kg h = 90m g = 9.8ms^-2

E = mgh
E = 800*9.8*90
= 7.056*10^5 J

Sirsh.
 
Physics news on Phys.org
Sirsh said:
Hi, i was just wondering with this question because I'm unsure if I am supposed to add both kinetic and gravitation energy into this answer? thanks.

1 A crane on a building site lifts an 8.00 × 102 kg load from ground level to a vertical height of 90.0 m at a constant speed of 2.00 m s–1. Ignore the mass of the cable.

A. How much work is done by the crane in lifting the load through this distance?
m = 800kg h = 90m g = 9.8ms^-2

E = mgh
E = 800*9.8*90
= 7.056*10^5 J

Sirsh.

Yes you also need to add the kinetic energy

Total work done = PE+KE = constant at all points in the motion (this is the law of conservation of mechanical energy)
 
Are you sure? supposedly it's W = F/s therefore W = mg/s, W = 800*9.8/90 ??
 
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