- #1
scluggy
- 2
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Two weights are hanging from a rope that goes through two pulleys as shown below:
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O |
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A B
A weighs 3 kg
B weighs 2 kg
B has an initial speed of 0.8 m/s
How far will B drop before A reaches a speed of 0,6 m/s?
The mass of the pulley and the cable can be neglected.
I have been trying to resolve this for some time now but will always get the wrong answer. The way I believe it should work is:
The potential energy + the kinetic energy for the system is the same at the start and end of this so we would have a conversion of energy from potential energy to kinetic energy. So I have set it up as having 0 potential energy to start off with (height = 0)
mgh = 2*9.81*0 = 0 and 3*9.81*0 = 0
The velocity for B would be twice that of the velocity for A due to the pulley system.
The initial kinetic energy would be:
0.5mv^2 = 0.5*2*0.8^2 and 0.5*3*-0.4^2
So the whole amount of energy for this would be
0.64-0.24 = 0.4
If we then set this equal to the same formula but for the A speed of 0.6 m/s we get
0.88 = (kinetic energy A) + (kineric energy B) + (potential energy A) + (potential energy B)
0.4 = 0.5*3*-0.6^2 + 0.5*2*1.2^2 + 3*9.81*h/2 + 2*9.81*h
Which makes h = - 100 / 6867
This is incorrect as the answer should be 0.224
Can someone see where I am going wrong and perhaps help with how this should be resolved? I have been trying to resolve this issue for some time now with different methods but will always come up with the wrong answer.
Thankful for any suggestions!
_______
| O
| / \
| | |
\ / |
O |
| |
A B
A weighs 3 kg
B weighs 2 kg
B has an initial speed of 0.8 m/s
How far will B drop before A reaches a speed of 0,6 m/s?
The mass of the pulley and the cable can be neglected.
I have been trying to resolve this for some time now but will always get the wrong answer. The way I believe it should work is:
The potential energy + the kinetic energy for the system is the same at the start and end of this so we would have a conversion of energy from potential energy to kinetic energy. So I have set it up as having 0 potential energy to start off with (height = 0)
mgh = 2*9.81*0 = 0 and 3*9.81*0 = 0
The velocity for B would be twice that of the velocity for A due to the pulley system.
The initial kinetic energy would be:
0.5mv^2 = 0.5*2*0.8^2 and 0.5*3*-0.4^2
So the whole amount of energy for this would be
0.64-0.24 = 0.4
If we then set this equal to the same formula but for the A speed of 0.6 m/s we get
0.88 = (kinetic energy A) + (kineric energy B) + (potential energy A) + (potential energy B)
0.4 = 0.5*3*-0.6^2 + 0.5*2*1.2^2 + 3*9.81*h/2 + 2*9.81*h
Which makes h = - 100 / 6867
This is incorrect as the answer should be 0.224
Can someone see where I am going wrong and perhaps help with how this should be resolved? I have been trying to resolve this issue for some time now with different methods but will always come up with the wrong answer.
Thankful for any suggestions!
