Question
Two masses are connected by a string. Mass A is on a flat roof and mass B is hanging over the edge of the roof. The mass on the roof is 15 kg and the hanging mass is 5 kg. The edge of the roof is 100 m above the ground. The two masses are initially at rest with the hanging mass 80 m above ground and the roof mass 50 m from the edge. There is no friction.
(Allowed to use only work/energy no kinematics)
after three seconds of motion the masses have moved 11 m.
a) calculate the speed of the masses at this instant
b) calculate tension in the rope
c) at this instant the mass on the roof travels along a section with friction where u= .577. how far along the roof does this mass move before stopping (the masses are still connected)
(Allowed to use only work/energy no kinematics)
after three seconds of motion the masses have moved 11 m.
a) calculate the speed of the masses at this instant
b) calculate tension in the rope
c) at this instant the mass on the roof travels along a section with friction where u= .577. how far along the roof does this mass move before stopping (the masses are still connected)
Answers
bobpursley
An atwood machine without a pulley.
work done by gravity= 11m*Bg
KE gained= 1/2 (A+B)v^2
set those two equal, solve for v
b) Tension in rope: on the second mass, Tension*11=1/2 A v^2, solve for tension
c) Now, you have initial KE of 1/2 (A+B)v^2, you are gaining GPE energy at Bg v rate, and friction is eating energy at mu*A*g*v rate
Now the average velocity during the slowing period is v/2, for a duration T seconds.
Frictionenergy=GPEgained+initialKE
mu*A*g*(v/2)T=mg(v/2)T+ 1/2 (A+B)v^2
solve for T, the time to slow to a stop.
how far did it travel? (v/2)T
work done by gravity= 11m*Bg
KE gained= 1/2 (A+B)v^2
set those two equal, solve for v
b) Tension in rope: on the second mass, Tension*11=1/2 A v^2, solve for tension
c) Now, you have initial KE of 1/2 (A+B)v^2, you are gaining GPE energy at Bg v rate, and friction is eating energy at mu*A*g*v rate
Now the average velocity during the slowing period is v/2, for a duration T seconds.
Frictionenergy=GPEgained+initialKE
mu*A*g*(v/2)T=mg(v/2)T+ 1/2 (A+B)v^2
solve for T, the time to slow to a stop.
how far did it travel? (v/2)T
thank you!
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