Block, Ramp, Friction, and Spring work done

AI Thread Summary
A man pulls a 20 kg block up a 33° incline for 3.5 m at a constant velocity, facing friction with a coefficient of 0.2. The work done by the man is calculated using the equations for gravitational and frictional work, leading to a total of 823.919 J, though initial calculations were incorrect due to misunderstanding the forces involved. The discussion clarifies that the normal force is not equal to the weight of the block, and adjustments in calculations for gravitational and frictional work are necessary. The correct approach involves using the sine and cosine of the incline angle to determine the work done accurately. The thread concludes with a successful resolution of the calculations after addressing the initial confusion.
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!
 
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