What are the equations needed to solve the elastic force on an incline problem?

AI Thread Summary
To solve the elastic force problem on an incline, key equations include those for kinetic energy, potential energy, and work done against friction. The speed of the package before it reaches the spring can be determined using forces acting along the incline, while the maximum compression of the spring involves equating kinetic energy to spring potential energy and work done against friction. The rebound height can be calculated using similar principles, considering energy conservation. The provided equation for compression, 0.5kx^2 = 4mgsin(53) - μ_k mg*cos(53)*4, is a crucial part of the solution. Understanding these relationships is essential for solving the problem effectively.
cycam
Messages
32
Reaction score
0

Homework Statement



A 2.00-kg package is released on a 53.1 ^\circ incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline View Figure . The coefficients of friction between the package and the incline are \mu_{s} \;=\; 0.40 and \mu_{k} \;=\; 0.20. The mass of the spring is negligible.

yf_Figure_7_41.jpg


1.] What is the speed of the package just before it reaches the spring?

2.] What is the maximum compression of the spring?

3.] The package rebounds back up the incline. How close does it get to its initial position?

Homework Equations





The Attempt at a Solution



I only have a limited amount of attempts to input the correct solution and I already have tried a couple different ways. I want to be guided in the right direction, not have the answer handed to me, please. Any help would be greatly appreciated. Thank you :smile:
 
Physics news on Phys.org
The forces that act on the block are friction, gravity. You can calculate the force along the incline and use the kinematics equations to find the velocity (if you have the force, you'll have the acceleration).

For the compression of the spring, you have the velocity, and hence the kinetic energy. That converts into the potential energy of the spring AND the work done against friction.

For the last part, again use the same considerations to find the max height of the block.

Btw, sin53=4/5 and cos53=3/5
 
argh. i can't seem to get the second part.
my equation is...
1/2kx^2 + (u_k)mgcos(53.1) - (1/2)(m)(v^2)=0
can anyone help?
 
Yeah,

The equation works out to be:
0.5kx^2=4mgsin(53) - \mu _k mg*cos53*4
 
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...
Back
Top