A crate of mass m1 on a frictionless inclined plane is attached to another crate of mass m2 by a massless rope. The rope passes over an ideal pulley so the mass m2 is suspended in air. The plane is inclined at an angle = 36.9°. Use conservation of energy to find how fast crate m2 is moving after m1 has traveled a distance of 1.4 m along the incline, starting from rest. The mass of m1 is 10.4 kg and the mass of m2 is 12.5 kg.
2007-06-21 04:27:01 · 3 answers · asked by terra_flare_aqua_ciel 1 in Science & Mathematics ➔ Physics
The crates are connected by a rope, so they each will move the same distance, same speed and same acceleration.
Also, conservation of energy can be used looking at the difference between potential energy and kinetic energy
Using g=9.81
the falling crate loses PE of
12.5*9.81*1.4
171.675 Joules
The incline angle I will assume is w/r/t the horizontal,
so in moving 1.4 m m1 gains PE of
10.4*1.4*9.81*sin(36.9)
85.76 Joules
Therefore the kinetic energy of the masses is
.5*(10.4+12.5)*v^2=
171.675-85.76
v^2=2*3.7517
v=2.74
I made an error in my previous post forgetting that
KE= 1/2*m*v^2.
I left off the 1/2
j
2007-06-21 07:29:11 · answer #1 · answered by odu83 7 · 0⤊ 1⤋
normal force is what rock exert on the inclined plane which is always m*g*cos(theta).and as cos(theta) varies from 0 to 1 so as force will vary from 0 to 2.
2016-05-21 11:52:25 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Why bother asking a question based on conditions that do noy exist, there is nothing to gain by it.
2007-06-25 04:18:34 · answer #3 · answered by johnandeileen2000 7 · 0⤊ 1⤋