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
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.
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
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