# Two objects connected by a cord, with a pulley.

1. Nov 4, 2009

### budd99

1. The problem statement, all variables and given/known data

The system shown (attached image) is released from rest and moves 50 cm in 1.0 s. What is the value of M? All surfaces are frictionless.

a. 0.42 kg
b. 0.34 kg
c. 0.50 kg
d. 0.59 kg
e. 0.68 kg

2. Relevant equations

Since the incline is 90 degress, I am trying to model this as an Atwood machine.

I don't understand how there can be any acceleration if the mass that we need to find is smaller than 3kg.

3. The attempt at a solution

moves 50cm in 1s., which means a is 0.5m/s2

So, using the formula
a = (m2 -m1)/(m1 + m2)g

0.5 = (m2 - 3)(3 + m2)9.80

m2 = 3.32

Which is not in the solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

File size:
4.1 KB
Views:
230
2. Nov 4, 2009

### Staff: Mentor

Not a good idea, since an Atwood machine has both ends of the string vertical.

Why do you think this?

No. Use kinematics to determine the acceleration.

Instead of using a canned formula (which happens not to apply in this case), derive your own from first principles. Apply Newton's 2nd law to each mass.

3. Nov 4, 2009

### budd99

Ok, I used 0.5 = 0 + 0.5a12 to find a, which gives 1.

So, if F = ma, then the F on the 3kg block is 3(1) which is 3.

On the block suspended in air, the F = mg.

Since we already found F = 3, then 3 = m(9.8), and finally m = .31

This still doesn't make sense, maybe I'm forgetting a T somewhere?

4. Nov 4, 2009

### Staff: Mentor

Exactly. Two forces act on the hanging mass.

(Also, from the diagram I can't read the mass of the sliding block. Is it 3 kg or 30 kg?)