Block, Ramp, Friction, and Spring work done

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a block being pulled up an incline with friction and a spring. The problem includes calculations for work done by the man pulling the block, the speed of the block when it reaches the horizontal surface, the spring constant, and the distance the block rebounds up the incline.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the man, questioning the correct formulation of work done against gravitational and frictional forces. There is exploration of the relationship between forces acting on the block and the correct interpretation of the equations involved.

Discussion Status

Some participants have provided insights into the calculations, while others are still questioning their assumptions regarding the forces and the setup of the problem. A participant claims to have resolved their confusion regarding the work done by gravity and friction, indicating some progress in understanding.

Contextual Notes

There are discussions about the definitions of work done by gravity and friction, as well as the normal force's relationship to the weight of the block. Participants are also addressing the implications of coordinate systems on their calculations.

Awwnutz
Messages
42
Reaction score
0
http://img412.imageshack.us/img412/7050/blockspringrampfrictionfy0.gif

A man pulls a block of mass m = 20 kg up an incline at a slow constant velocity for a distance of d = 3.5 m. The incline makes an angle q = 33° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.2.

a) What is the work Wm done by the man?

b) What is the speed v of the block when it first reaches the horizontal surface?

c) What is the spring constant k of the spring?

d) How far up the incline d1 does the block rebound?



Relevant equations:
Wtotal = Change in Kinetic energy
Ffriction = Coefficient of friction(Fnormal)
Fspring = kx
(1/2)mv^2


I know there are the force of friction, the man, and gravity on the box. I started by saying:
Wman - Wgrav - Wfriction = Change in KE
Wgrav = m*g(in x-direction)*(Change in height) -->(20kg)(9.81m/s^2*sin33)(3.5sin33m)
Wfriction = (coefficient of friction)*m*g*distance -->(.2)(20kg)(9.81m/s^2)(3.5m)
Change in KE = 0
Wman = 823.919J, but this is not right...what am i doing wrong?
 
Last edited by a moderator:
Physics news on Phys.org
What am i doing wrong with work done by the man? Isn't it just
Wman-Wgrav-Wfriction = 0
Wgrav = -m*g*h
Wfric = -(Coefficient of friction)(m*g)(distance)
 
Awwnutz said:
What am i doing wrong with work done by the man? Isn't it just
Wman-Wgrav-Wfriction = 0
Wgrav = -m*g*h
Wfric = -(Coefficient of friction)(m*g)(distance)
The work done by gravity is either -mgh or -mg(sin theta)*d. You used W_grav = -mg(sin theta)*d(sin theta) .

Also, when calculating the friction force, the normal force and the weight are not the same.
 
ok so your saying

Wgrav = -(20kg)(9.81m/s^2)(sin33)(1.9m)

And if your turn your coordinate axis so the normal force is going in the positive y direction wouldn't that make the normal force equal and opposite of the weight of the block?
 
i figured it out

Wgrav = m*g*(sin theta)*d
Wfric = (Coefficient of friction)*(m*g*cos theta)*d

after that it was quite easy to figure out the rest.

Thanks PhanthomJay!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
18
Views
3K
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
17
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
20
Views
3K