Two blocks and a pulley homework

AI Thread Summary
The problem involves two blocks connected by a string over a frictionless pulley, with one block on a rough surface and the other hanging. The 3 kg block experiences kinetic friction, while the 7 kg block falls a distance h before stopping. The discussion emphasizes using energy conservation principles to solve for the falling distance and the kinetic energy of both blocks at a specific descent of 10.5056 m. Participants are working through equations involving gravitational potential energy, spring potential energy, and kinetic energy, while also accounting for work done by friction. The conversation focuses on setting up the problem correctly to apply these concepts effectively.
duketogo
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Homework Statement



The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 3 kg block lies
on a rough horizontal surface with a constant coefficient of kinetic friction 0.1. This block
is connected to a spring with spring constant 5 N/m. The second block has a mass of 7 kg.
The system is released from rest when the
spring is unstretched, and the 7 kg block falls
a distance h before it reaches the lowest point. acceleration of gravity is 9.8m/s/s
Calculate the falling distance h where the
7 kg blocks stops. Answer in units of m.
Note: When the 7 kg block is at the lowest point velocity is zero

and then
calculate the moment when the 7kg block has descended by a distance of 10.5056m, where 10.5056 is less than h. at this moment calc the sum of kinetic energy for the two blocks in joules.


Homework Equations


E=mgh= M m work
(1/2mv^+1/2kx) + (1/2mv^2+ mg(h-x) + (f sub k * x)
this equaition was given by a friend but I can't seem to make it work
h=2(m-uM)g/k
any help on setting up the problem would be appreciated

for the second part here's what I got thus far total k.e.= 1/2mv^2 +1/2Mv^2= mgx - 1/2kx^2-uMgx
but I'm not sure about it.

there's a picture of the diagram here

ac8pyf.jpg
 
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duketogo said:

Homework Equations


E=mgh= M m work
(1/2mv^+1/2kx) + (1/2mv^2+ mg(h-x) + (f sub k * x)
this equaition was given by a friend but I can't seem to make it work
h=2(m-uM)g/k
any help on setting up the problem would be appreciated
The basic idea is energy conservation:
Mechanical Energy(initial) + Work done by friction = Mechanical Energy(final)

Mechanical energy is the sum of spring PE (½kx²), gravitational PE, and KE.
Work done by friction (which is negative) = -μNx
 
Doc Al said:
The basic idea is energy conservation:
Mechanical Energy(initial) + Work done by friction = Mechanical Energy(final)

Mechanical energy is the sum of spring PE (½kx²), gravitational PE, and KE.
Work done by friction (which is negative) = -μNx


okay I'll try it out
 
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