Calculating Velocity: Roofer's Toolbox Sliding on a 36deg Slanted Roof

  • Thread starter Thread starter rezal
  • Start date Start date
AI Thread Summary
The discussion focuses on a physics problem involving a toolbox sliding down a slanted roof at a 36-degree angle. The user seeks validation for their calculations, which involve energy transformations and the effects of friction. Feedback emphasizes the importance of understanding the problem-solving process, including writing out reasoning and energy equations before substituting numbers. It is suggested that the user revisit their energy calculations, as they may have misapplied the friction force and the weight of the toolbox. Gaining confidence in problem-solving requires a solid grasp of the underlying concepts and careful attention to detail.
rezal
Messages
5
Reaction score
0
Hello, I'm practicing by doing the problems in my textbook but I have no way of knowing if my answers to positive questions are correct or not so can someone tell me if I did this problem correctly?

1. Homework Statement

While a roofer is working on a roof that slants at 36deg above the horizontal, he accidentally nudges his 85.0-N toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Homework Equations


KE + U + W = KE + U sin(theta)=o/4.25

The Attempt at a Solution


U + W = KE o=4.25sin36=2.49
mgcos(theta)(h)+wsin(theta)(u_k)=KE
(85*cos(36)*2.49)+(85*sin(36)*22)=KE
KE=1270.38
v=sqrt((2*KE)/m)
V=17.12 m/s?
 
Physics news on Phys.org
Welcome to PF;
If you don't know if the answer is correct or not, then you did not understand the work. You are, after all, training to be able to solve problems where nobody knows the answers so the sooner you start figuring out how to gain confidence in your own work the better.

The way to gain confidence is to write out your reasoning, and reality check the results. Here you write out, in words, what the energy transformations are.
Then write down an energy equation that says the same thing.
Then go through the factors in the energy equation and write them out in terms that you are given ...

It is also best practice to sub numbers in at the very end if at all possible - do the algebra first.

Off the first glance, you should take another look at the energy calculations ... 22N is the friction force, so you don't need the coefficient of kinetic friction to calculate it. You are given the weight, so you don't need mg in your equation. the distance to the edge reads to me like the distance along the roof ... stuff like that.
 
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...
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...
Back
Top