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posted by  Danyal on 11/1/2009 11:48:37 PM  |  status: Closed  |  Earned Karma: 67

Pulley and Energy (Lifesaver Rating)

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N/A N/A N/A N/A 11/2/2009 at 1:00:00 PM
Question Details:


A 20 kg mass A is suspended by a pulley to a cable attached on the ceiling at one end, and a spring at its other end. The mass is to be dropped delicately to the floor from an initial position at which the bottom of the pulley is at the same elevation as the fixed end of the cable, 4 m above the floor (so
that the upper length of cable is then straight and horizontal). Initially, the spring is unstretched and has a length of 1 m.
a. What spring stiffness is required to ensure that mass A reaches the
ground C with zero velocity?
b. What is then the velocity of A when it is 0.5 m above the ground?

Danyal Ali
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Guru
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posted by Sapere Aude on 11/2/2009 8:20:56 AM  |  status: Live
Asker's Rating: Not Helpful   
Response Details:
a) when A has no velocity at the bottom it means that all potential energy of A is conducted to the spring:
m*g*h = 0.5*k*x^2
where m=20, h=3,g=9.8, x=3
solve for k
k=11.4 N/m

 
b) so its height is 0.5 m, at that position it has a velocity of v, lets use conservation of energy
initially there is only potential energy = mgh
when mass is at0.5 m then there is again a potential energy but smaller, a kinetic energy and also energy stored in the spring. They are totally:
m*g*h1 + 0.5*m*v^2 + 0.5*k*x^2 = m*g*h
20*9.8*0.5 +0.5*20*v^2 +0.5*11.4*2.5^2 = 20*9.8*3
solve for v= 6.74 m/s
enjoy
Have Fun!
Guru
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posted by Root_Locus(MNK) on 11/2/2009 9:23:31 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:


a)initially spring cable length is 8m + 4m = 12m
  at the final position length = 2*√(32+42) + 4m = 14m  {pulley goes down by 3m and it goes in the mid way of  8m                                                                                                horizontal distance}
so the extension in the spring = 2m
Now, according to question final velocity of block should be zero.
It means all the potential energy lost by the mass should be converted to spring potential energy.
applying conservation of energy
(1/2)k*22 = mgh = 20*9.81*3   {mass goes down by 3m only}
or, 2k = 20*29.43
or, k = 294.3 N/m 

b)when mass is 0.5m above the ground
spring cable length = 2*√(2.52+42) + 4m = 13.434m
so the extension in the spring = 1.434m
mass goes down from its initial position by 2.5m
again applying conservation of energy
loss in P.E. = Gain in K.E. + spring energy
20*9.81*2.5 = (1/2)*20*v2 + (1/2)*294.3*1.4342
490.5 = 10v2 + 302.593
or, 10v2 = 187.907
or, v = √18.7907 = 4.334 m/s
Best Wishes
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