Phy 231
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A mass of .54 kg rests on a frictionless tabletop, attached by a string running horizontally to and then over a pulley to a mass of .1134 kg.
• When the system is released what will be its acceleration?
• What is the tension in the strings?
Wt = mg
Wt = (.1134 kg) *(9.8) = 1.11 N
F = ma
Note that m is the total mass .1134 kg+ .54 kg, from which your next step follows:
1.11 N = (.1134 kg + .54 kg ) * a
a = 1.11 N / (.6532 kg) = 1.7 m/s^2
The system will have an acceleration of 1.7 m/s^2.
Tension = Wt
The tension in the string will have a force of 1.11 N.
Each mass accelerates at 1.7 m/s^2.
The suspended mass experiences a downward force of 1.11 N. If the tension in the string was 1.11 N, then it would be in equilibrium, not accelerating at 1.7 m/s^2. So your tension is not correct.
What is the net force on the suspended mass?
You know the gravitational force on the mass, so in order to achieve the required net force, what must be the tension?
You also know that the mass on the tabletop accelerates at 1.7 m/s^2, and that net force on this mass is the tension.
What therefore is the tension?
Do your two results for the tension agree?
A block of mass 1.5 Kg is held stationary on a level frictionless tabletop. A mass of .22 Kg is attached to the block by a string over which runs horizontally from the block to a pulley located at the edge of the table; the mass hangs freely from the string over the pulley.
• As the hanging mass descends 1.5 meters from rest the block is pulled 1.5 meters along the tabletop (it's a good-sized table--and pretty high too). How much work does gravity do on each mass during the process?
• Using the definition of KE, determine the velocity of the system after having traveled the 1.5 meters.
F = 1.5 kg *9.8 m/s^2 = 14.7 N
W = 14.7 N * -1.5 m = -22.05 J
If the mass on the tabltop descended 1.5 m that would be correct, but it doesn't descend. It's on a level tabletop.
F = .22*9.8 = 2.156 N
W = 2.156N * -1.5 = -3.234 J
Gravity does -22.05 Joules of work on the first mass and -3.234 Joules of work on the second mass.
K = 1/2 mv^2
Total Work = -22.05 + -3.234 = 25.284 = Kinetic Energy
25.284 = 1/2 * (1.5+.22) * v^2
29.4 = v^2
v = 5.4 m/s
If both masses were descending without interference this would be correct. This result should therefore be identical to what you would get for a free fall of 1.5 m from rest.
However this doesn't correspond to the given system, which has one mass sliding along a tabletop.
Had you set the 3.234 J of work equal to the KE of the system, you would have obtained (solving 1/2 m v^2 = KE for v)
v = +-sqrt( 2 * KE / m) = +- sqrt( 2 * 3.234 J / (1.72 kg) ) = +- 1.9 m/s, approx.
Depending on your chosen positive direction, v will be positive or negative, as fits the situation.
The velocity of the system is 5.4 m/s after traveling 1.5 meters.
** **
I'm still not 100% sure about pulley problems so I was wondering if you could check my work. These are some problems from one of the tests.
** **
Not bad. You need to modify a couple of ideas, but I think you'll understand.
&#Let me know if you have questions. &#