Question

Two objects with masses of 2.45 kg and 4.15 kg are connected by a light string that passes over a light frictionless pulley to form an Atwood machine, as in Figure 5.14a. Figure 5.14a (a) Determine the tension in the string. Incorrect: Your answer is incorrect. N (b) Determine the acceleration of each object. m/s2 upwards (2.45 kg mass) m/s2 downwards (4.15 kg mass) (c) Determine the distance each object will move in the first second of motion if they start from rest. m (2.45 kg mass) m (4.15 kg mass)

Answer 1

I shall assume that the 4.15 kg mass is going to go down.Let us begin by focusing on the 2.45 kg mass
Forces up - forces down = ma
T - 2.45g = 2.45a (The a is of course common, as is the T, as they are part of the same system)
Now for the 4.15 mass,   4.15g - T = 4.15a
Now we add those two equations to get, 4.15g - 2.45g = 2.45a + 4.15a
Which should equal   1.7g = 6.6a
So a = (1.7g)/(6.6) = 2.52 m/s^2 Which is actually the answer to the (b) part.

We can now use this value of a in either of the equations to find a value for T. I picked the first one.
2.45g - T = 2.45(1.7g/6.6)
T = 2.45g(1 - (1.7/6.6)) = 17.83 N

For (c), we shall use the equation s=ut + 1/2at^2
s = what we're looking for
u = 0
t = 1
a = (1.7g)/(6.6) (it could be negative depending on which block you're focusing on, but they both move the same distance so it shouldn't matter...)
So, s = 0 + (1/2)(1.7g/6.6)(1) =1.26 m