Solving the Problem of a Bus Running Out of Fuel on a Hill

Thread starter BuBbLeS01
Start date Oct 24, 2007
Tags

Bus Fuel Hill Running

AI Thread Summary

To determine the speed required for a bus to coast to the top of a 3.54 m hill after running out of fuel, the key equation used is the conservation of energy: 1/2mv^2 = mgh. The mass of the bus cancels out, simplifying the equation to v^2 = 2gh. Substituting the height of the hill into the equation allows for the calculation of the necessary speed. The discussion emphasizes the need for clarity on how to approach the problem despite limited initial information. Understanding energy conservation is crucial for solving this type of physics problem.

Oct 24, 2007

BuBbLeS01

Homework Statement

A bus runs out of fuel as it approaches a hill. If the hill is 3.54 m high, how fast must the bus be traveling in order to coast just to the top of the hill?

Homework Equations

The Attempt at a Solution

How do we solve this problem?? We aren't given any
information (mass, forces or anything)?

Physics news on Phys.org

Oct 24, 2007

logglypop

1/2mv^2 = mgh

Oct 24, 2007

BuBbLeS01

thank you much

Thread 'Errors and significant figures'

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...

View full post »

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...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »