Block of ice on frictionless surface

Thread starter rebeccc
Start date Sep 15, 2008
Tags

Block Frictionless Frictionless surface Ice Surface

AI Thread Summary

A 1.6-meter rope pulls a 9.0 kg block of ice across a frictionless surface, resulting in an acceleration of 1.0 m/s². To determine the force pulling the block of ice, Newton's second law (F=ma) can be applied, yielding a force of 9.0 N. The force exerted on the rope is equal to the force acting on the block, as both are connected and moving together. Participants in the discussion emphasize the importance of identifying the forces at play and using appropriate equations and units. Understanding these concepts is crucial for solving the problem effectively.

Sep 15, 2008

rebeccc

A 1.6 -long, 550 rope pulls a 9.0 block of ice across a horizontal, frictionless surface. The block accelerates at 1.0 .

1) How much force pulls forward on the block of ice?

2) How much force pulls forward on the rope?

Physics news on Phys.org

Sep 15, 2008

mgb_phys

Science Advisor

Homework Helper

Welcome to PF - please have an attempt at the question.
What equations do you know, what are the units?
What forces are acting ?

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 »