Block on a Pulley: Determine Speed 2s Later

  • Thread starter Thread starter KillerZ
  • Start date Start date
  • Tags Tags
    Block Pulley
AI Thread Summary
The discussion revolves around calculating the speed of a 10-lb block moving downward after 2 seconds, considering the dynamics of two blocks connected by a pulley. The initial speed of block A is 6 ft/s, with block B experiencing kinetic friction. Participants analyze forces acting on both blocks, derive equations for tension and acceleration, and correct errors in calculations. The final consensus indicates that the speed of block A after 2 seconds is approximately 34.73 ft/s, following adjustments to the equations and ensuring consistent signs for acceleration. The importance of precision in calculations and the correct interpretation of tension in the system is emphasized throughout the discussion.
KillerZ
Messages
116
Reaction score
0

Homework Statement



At a given instant the 10-lb block A is moving downward with a speed of 6 ft/s. determine its speed 2s later. Block B has a weight of 4-lb, and the coefficient of kinetic friction between it and the horizontal plane is \o_{k} = 0.2. Neglect the mass of the pulleys and cord.

2uo0t91.png


Homework Equations



v = v_{0} + a_{A}t

The Attempt at a Solution



FBD:
2h6zltw.png


Block A:

+\downarrow \sum F_{y} = ma_{y}

10 - 2T = (0.311)(a_{A})

Block B:

+\rightarrow \sum F_{x} = ma_{x}

T - (0.2)(4) = (0.124)(a_{B})

+\downarrow \sum F_{y} = ma_{y}

4 - N = 0

Kinematics:

2S_{A} + S_{B} = l

2a_{A} + a_{B} = 0

2a_{A} = -a_{B}

Solving for T:

10 - 2T = (0.311)(a_{A})

10 - 2T = -(0.311)(a_{B})

a_{B} = -\frac{10 - 2T}{0.311} = \frac{T - (0.2)(4)}{0.124}

T = -16lb/2 = -8lb

a_{B} = -70.84ft/s^{2}

a_{A} = 83.72ft/s^{2}

v = v_{0} + a_{A}t

v = 6 + (83.72)(2) = 173.44 ft/s I am not sure if this is right because that seems fast.
 
Physics news on Phys.org
KillerZ said:
2a_{A} = -a_{B}

Solving for T:

10 - 2T = (0.311)(a_{A})

10 - 2T = -(0.311)(a_{B})

you forgot a factor (1/2) here.

a_{B} = -\frac{10 - 2T}{0.311} = \frac{T - (0.2)(4)}{0.124}

T = -16lb/2 = -8lb

T = -8 lb is not a solution of the above equaton. Try to give some more detail when you
try to solve for T again.
The negative tension in a rope should have tipped you off that something was wrong.
 
I didn't have a chance to get back to this but I finally solved it I think:

2a_{A} = -a_{B}

a_{A} = -\frac{a_{B}}{2}

10 - 2T = (0.311)(a_{A})

10 - 2T = -(0.311)(\frac{a_{B}}{2})

a_{B} = -\frac{20 - 4T}{0.311} = \frac{T - (0.2)(4)}{0.124}

-2.48 + 0.50T = 0.31T - 0.25

T = 2.769 lb

a_{A} = 14.37ft/s^{2}

a_{B} = 15.85ft/s^{2}

v = v_{0} + a_{A}t

v = 6 + (14.37)(2) = 34.73 ft/s
 
KillerZ said:
I didn't have a chance to get back to this but I finally solved it I think:

2a_{A} = -a_{B}

I'm sorry, this was an error I missed. a_A and a_B should have the same sign.

-2.48 + 0.50T = 0.31T - 0.25

T = 2.769 lb

this T is not the solution of the last equation. I get 0.19 T = 2.23, so T = 11.7
your also using only 2 digits of precision here, and then give the final answer with much
more digits.
 
10 - 2T = 0.311(aA)
T - 0.8 = 0.124(aB)
aA = 2*aB
So 10 - 2T = 0.622*aB
T - 0.8 = 0.124*aB
So ( 10 - 2T)/(T - 0.8) = 0.622/0.124.
Now solve for T.
 
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