Work required for truck up a hill problem

  • Thread starter Thread starter gigglin_horse
  • Start date Start date
  • Tags Tags
    Hill Truck Work
AI Thread Summary
More work is required to bring a fully loaded truck up a hill compared to a lightly loaded truck due to the difference in kinetic energy (KE). The initial misunderstanding involved equating work to force multiplied by time, which is incorrect as work is defined as force multiplied by distance. The discussion clarifies that while both scenarios can yield the same speed, the heavier truck requires more work due to its greater mass. The relationship between work and change in kinetic energy is confirmed, supporting the conclusion that a greater load necessitates more work. Ultimately, the heavier truck indeed requires more work to reach the same speed.
gigglin_horse
Messages
32
Reaction score
0

Homework Statement



"Is more work required to bring a fully loaded truck up to a given speed than the same truck lightly loaded? Defend your answer"

Homework Equations



I put:
Work = Force x distance
Distance = speed x time
As speed in a common factor, we take it out
Work = Force x time....Is this correct so far?

So the work is the same.
Small force, long time, or short time, big force.


But doesn't work = change in KE?
so W = 1/2 m x v^2
...So the greater one does require more work?

I don't know...

Help me please!


The Attempt at a Solution

 
Physics news on Phys.org


gigglin_horse said:
I put:
Work = Force x distance
Distance = speed x time
As speed in a common factor, we take it out
Work = Force x time....Is this correct so far?
No, not correct. (If work = force*distance, which it does, how can it also equal force*time? The units won't even make sense. FYI: force*time = change in momentum, not work.)

So the work is the same.
Small force, long time, or short time, big force.
No.

But doesn't work = change in KE?
Yes!
so W = 1/2 m x v^2
...So the greater one does require more work?
Yes!
 


Brilliant, thank you
=]
 
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...

Similar threads

Back
Top