Work, energy and power on a slope.

In summary, the conversation discusses a diagram showing two particles, A and B, where A moves on a rough inclined plane and B hangs freely attached to A. The coefficient of friction between A and the plane is 1/4 and B is initially hanging 3m above the ground. When the system is released, A moves up the plane. The initial acceleration of A is found using the equations F=ma and K.e. = 1/2mv2. To calculate the total distance moved by A before it first comes to rest, the principle of conservation of energy is used. The kinetic energy of A at the moment the string breaks is also calculated using the work of friction.
  • #1
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Homework Statement


The diagram shows a particle A of mass m which can move on the rough surface of a plane inclined at an angle theta to horizontal ground, where theta = arcsin 0.6. A second particle B of mass 2m hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at D. The other end of the string is attached to A. The coefficient of friction between A and the plane is 1/4. B is initially hanging 3m above the ground and A is 2m from D. When the system is released from rest with the string taut A moves up the plane.
(a) Find the initial acceleration of A.
When B has descended 1m, the string breaks. By using the principle of conservation of energy.
(b) calculate the total distance moved by A before it first comes to rest.

http://img135.imageshack.us/img135/9503/scan0001small.gif [Broken]

Homework Equations


F=ma
K.e. = 1/2mv2

G.p.e = mgh

The Attempt at a Solution


a) 2mg - T = 2ma
T - cos36.87 = ma
2mg - 0.8mg = 3ma
1.2g = 3a
a = 3.92ms-2

b) mgh = 2mg
2(0.6) = 1.2

I got part a right but b is meant to be 1.5m. Not sure why I ain't getting the right answer...
 
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  • #2
For b) determine the kinetic energy of A at the moment the string breaks.

The work of friction .8m*g*X = 1/2m*v2 is what will slow it to a stop.

X is the additional distance past 1m that you are looking for.
 
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  • #3
Wow, thanks again for your help! Got the right answer now! Thank you!
 

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