Gravitational ep in a uniform field

Thread starter kelly242
Start date May 3, 2011
Tags

Field Gravitational Uniform Uniform field

AI Thread Summary

The problem involves calculating the change in gravitational potential energy for a 10.0 kg box sliding down a 15.0 m ramp inclined at 20.0 degrees, with a friction force of 40.0 N. The initial attempt calculated work done against friction as 563.8 J, but the correct change in gravitational potential energy is -503 J. To find this, the formula V = mgh should be used, requiring the initial height to be determined. The discussion emphasizes the importance of correctly applying gravitational potential energy concepts in physics problems. Understanding the relationship between work, energy, and friction is crucial for accurate calculations.

May 3, 2011

kelly242

Homework Statement

A 10.0 kg box is slid down15.0 m along a 20.0o ramp. The force of friction is 40.0 N. What is the change in the gravitational potential energy of the box?

Homework Equations

W=Fdcos of theta

The Attempt at a Solution

W=(40 N)(15 m)(cos 20)
=563.8J
The actual answer is -503 J...

Physics news on Phys.org

May 3, 2011

TobeHode

Try:

V = mgh

Then just figure out what the initial height is.

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

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 »

Thread 'A cylinder connected to a hanged 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 »