Solve Work on Inclined Plane for m1, a, h, m2, uk

AI Thread Summary
The discussion revolves around calculating the work done on two blocks connected by a pulley system, with one block on an inclined plane and the other hanging. Key calculations include the work done by gravitational forces, tension, and friction for both blocks as they move a distance h. The total work on the system is derived from these individual contributions, leading to an expression for the final speed of the blocks. Additionally, there is a secondary problem involving three blocks connected in a similar manner, where the goal is to determine the coefficient of kinetic friction for constant speed. The conversation also touches on the importance of clear communication in forum posts.
oreosama
Messages
51
Reaction score
0

Homework Statement


A block of mass m1 is placed on an inclined plane with slope angle a and is connected to a second hanging block of mass m2 by a cord passing over a small frictionless pulley the coefficient of kinetic friction is uk the system is released from rest with block m2 dropping a distance h.

given m1, a, h, m2, uk
calc the work due to each of the forces on block 1 in moving h
calc the work due to each of the forces on block 2 in moving h
calc the total work on the sys and final speed of the blocks

Homework Equations



w = F*d

The Attempt at a Solution



http://i.imgur.com/IxtPw.png

for block1

Wg = -h*mg*sin a

Wt1 = hT

Wn=0

Wff = -h*uk*mg*cos a

Wtot = hT - *mg*sin a - h*uk*mg*cos a


for block2

Wg = mgh

Wt = -hT

Wtot = mgh - hT


hT - *mg*sin a - h*uk*mg*cos a + mgh - hT = 1/2*m*Vf^2 + 1/2*m2*Vf^2 - 0 (start from rest)

sqrt( 2/(m1+m2) * (mg*sin a - h*uk*mg*cos a + mgh)) = Vf



----

blocks a, b, c are connected as shown. block a,b,c have same mass m and coefficient of kinetic friction between each block and the surface is uk. block c descends with constant velocity. use 30 for angle of incline. given m h uk determine

work due to forces on A moving h
work due to forces on B moving h
work due to forces on C moving h
solve for uk in terms of the other givens for the system to move at constant speed



http://i.imgur.com/DtwxJ.png


block A

Wff= -uk*mg*h

Wt= T1*h

block B

Wt2 = T2*h
Wg = -mg*sin a*h
Wff= -uk*mg*cos a * h
Wt1 = -T1*h

wtot = T2*h - mg*sin a * h - uk*mg*cos a * h - T1* h

block C

Wg = mg*h
Wt2 = -T2*h

Wtot = mg*h - T2*h


T1*h - uk*mg*h + T2*h - mg*sin a * h - uk* mg* cos a * h - T1*h + mg*h - T2*h = 0(constant speed implies vf v0 are the same meaning no change in kinetic energy right??)

-uk*mg*h - uk*mg*cos a * h = mg*sin a * h - mg*h

-uk - uk*cos a = sin a - 1

-uk(1 + cos a) = sin a - 1
uk = - (sin a - 1/ (1+cos a))

uk = 1/2 / (2+sqrt(3))/2

uk = 1/(2+sqrt(3))






want to be sure I am not doing anything horribly wrong thanks for any help.
 
Physics news on Phys.org
oreosama said:

Homework Statement


A block of mass m1 is placed on an inclined plane with slope angle θ and is connected to a second hanging block of mass m2 by a cord passing over a small frictionless pulley. The coefficient of kinetic friction is μk. The system is released from rest with block m2 dropping a distance h.

given m1, θ, h, m2, μk
calc the work due [STRIKE]to[/STRIKE] by each of the forces on block 1 in moving h
calc the work due to each of the forces on block 2 in moving h
calc the total work on the sys and final speed of the blocks

http://i.imgur.com/IxtPw.png
...

----

blocks a, b, c are connected as shown. block a,b,c have same mass m and coefficient of kinetic friction between each block and the surface is uk. block c descends with constant velocity. use 30 for angle of incline. given m h uk determine

work due to forces on A moving h
work due to forces on B moving h
work due to forces on C moving h
solve for uk in terms of the other givens for the system to move at constant speed

http://i.imgur.com/DtwxJ.png
...

want to be sure I am not doing anything horribly wrong thanks for any help.

Wow! A little bit of punctuation, etc. sure helps ! ...along with a few other changes.
 
didnt seem to be an issue on other posts, sorry
 
oreosama said:
didnt seem to be an issue on other posts, sorry
Do you mean, in your other posts?

In many of those you did, at least, use periods.

You probably should take a bit of time to read rules for posting in this Forum. Here are two short excerpts.
General Posting Guidelines
...Pay reasonable attention to written English communication standards. This includes the use of proper grammatical structure, punctuation, capitalization, and spelling. ...

Do not hijack an existing thread with off-topic comments or questions--start a new thread.
 
are you having trouble interpreting what I've typed? clearly not or you wouldn't be capable of correcting it by replacing one word and telling me to use more periods. I think that fulfills reasonable english communication.

now read the last sentence of your post. do you not see the irony?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top